if $x=y$ we get $2f(xf(z)+x)=(z+1)f(2x) \Rightarrow f(xf(z)+x)=\frac{z+1}{2} \cdot f(2x)$ therefore if $a > \frac{1}{2}$ and $b\in Im(f)$ then $ab \in Im(f)$, Proving that $f$ is surjective. let $a$ be a number s.t $f(a) = 1$, then $(a+1)f(x+y) = f(xf(a)+y)+f(yf(a)+x)= 2f(x+y) \Rightarrow a=1$, so $f(1) = 1$ then $z+1 = (z+1)f(\frac{1}{2} + \frac{1}{2})=2f(\frac{f(z)+1}{2}) \Rightarrow f(\frac{f(z)+1}{2}) = \frac{z+1}{2}$ substituting $z=\frac{z+1}{2}$ we get $f(\frac{z+3}{4})=\frac{z+3}{4}$, so for $x>\frac{3}{4}, f(x)=x$. let $z$ be any number, and $x,y$ sufficiently big numbers so $x+y,xf(z)+y,yf(z)+x > \frac{3}{4}$, then: $(z+1)(x+y)=(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x)=xf(z)+y+yf(z)+x =(f(z)+1)(x+y) \Rightarrow f(z) = z$

Basically what I submitted on the actual APMO.

i found. Quite similar as above

$\boxed{Solution}$ $(z+1)f(a+b)=f(a\cdot f(z)+b)+f(b\cdot f(z)+a)$ $i)x\cdot f(z)+y=M$ $i)a\cdot f(z)+b=K$ $ii)y\cdot f(z)+x=N$ $ii)b\cdot f(z)+a=L$ We know, $M+N=K+L$ $x+y=a+b$ Then.... $x\cdot f(z)=M-y$ and $a\cdot f(z)=K-b$ Then: $f(z)(x-a)=M+b-K-y$ $\boxed{Case1:}$ $a=b$ and $x=y$ Then. $x\cdot (f(z)+1)=M$ $a\cdot (f(z)+1)=K$ Then $M+K=(f(z)+1)(a+x)$ $\boxed{Case2:}$ $b=a$ and $y=x$ then $y\cdot (f(z)+1)=N$ $b\cdot (f(z)+1)=L$

Lengthy solution incoming... too bad i didnt found it during the contest I will edit the latex later.. (Edit: latex editted) $x=y$, $[(z+1)/2]*f(2x)=f(xf(z)+x)$. say $f(u)=f(v), z=u, z=v$, then $(u+1)f(x+y)=(v+1)f(x+y)$=> $u=v$ since $f(x+y)>0$. So f injective..........(1) $x=y, (z+1)f(2x)=2f(xf(z)+x)$ => $[(z+1)/2]f(2x)=f(x(f(z)+1))$..........(2) $x=1, [(z+1)/2]f(2)=f(f(z)+1)$ => f is surjective on $(1/2*f(2), +\infty)$. For any fixed $z<1$, take any $t \in (1/2*f(2), +\infty)$, take $a$ s.t $f(2a)=t$. Then $[(z+1)/2]t=[(z+1)/2]f(2a)=f(xf(a)+x)$ => $[(z+1)/2]t$ is also in $Im(f)$. So $t \in Im(f)$ => $[(z+1)/2]t \in Im(f)$ => ... => $[[(z+1)/2]^k]t \in Im(f)$. But $z<1$ then $[[(z+1)/2]^k]t$ can be arbitarily small, but it is in $Im(f)$. So f is surjective on postive reals, ie f is bijective..........(3) Suppose there is an $b\in R+$ so that $f(b)=1$. Take $z=b, x=y$, then $(b+1)f(2x)=2f(xf(a)+x)=2f(x)$ => $b=1$, since $f(2x)>0$. But such $a$ must also exist by bijectivity. So $f(a)=1 \iff a=1$, in particular $f(1)=1$..........(4) Look at (2), take $x=\frac{1}{2}$ and $x=\frac{1}{(f(z)+1)}$ at (2). We have $f(\frac{2}{[f(z)+1]})=\frac{2}{(z+1)}$ and $f(\frac{[f(z)+1]}{2})=\frac{(z+1)}{2}$. $f(z)+\frac{1}{2}$ can take any value from $(\frac{1}{2},+\infty)$. So for any $w \in (\frac{1}{2},+\infty)$, exist $z$ s.t $\frac{[f(z)+1]}{2}=w$ So this gives $f(1/w)=2/(z+1)=1/[(z+1)/2]=1/f(w)$ for all $w \in (\frac{1}{2},+\infty)$ If now $x<\frac{1}{2}$, then we can take $w=1/x$, then $f(1/x)f(x)=1$, same equation. So this means $f(x)f(1/x)=1$ for all $x\in R+$..........(5) In (2) again, $x=f(t), z=1/t$, then $(t+1)f(2f(t))=2tf(f(t)f(1/t)+f(t))$ By (5), $(t+1)f(2f(t))=2tf(f(t)+1)=t*(t+1)f(t)$ => $f(2f(t))=tf(2)$.........(6) For any $p \in (\frac{1}{2}, +\infty)$ we can choose $z$ s.t $p=[f(z)+1]/2$. Then $f(2p)=f(f(z)+1)=[(z+1)/2]f(2)=f(2)[(z+1)/2]=f(2)f([f(z)+1]/2)=f(2)f(p)$. So $f(2p)=f(2)f(p)$ for all $p \in (\frac{1}{2}, +\infty)$. So in fact $f(2p)=f(2)f(p)$ for all $p \in R+$..........(7) (By using similar idea like (5).) So $xf(2)=f(2f(x))=f(f(x))f(2)$ => $f(f(x))=x$ for all $x \in R+$..........(8) Now we replace (*) by taking $x=a, y=b, z=f(c)$, we ignore $x,y,z$ from now. We get $[f(c)+1]f(a+b)=f(ac+b)+f(bc+a)$ for all $a, b, c \in R+$..........(**) Take $a=b=1, [f(c)+1]f(2)=2f(c+1)$..........(9) Let $f(2)=k$, then by (7), $f(4)=f(2)^2=k^2$. But we use (9) on $c=2$, then $a(a+1)=2f(3)$ => $f(3)=a(a+1)/2$. And (9) on $c=3$, then $a^2=f(4)=a[a(a+1)/2+1]/2$ => $2a=[a(a+1)/2]+1$. Solve, then we get $4a-2=a(a+1)$ => $a^2-3a+2=0$ => $a=1$ or $a=2$. But if $a=1$, then $f(2)=1=f(1)$, contradiction with (3). So $a=2$, so $f(2)=2$............(10) So (9) becomes $f(z)+1=f(z+1)$ for all $z \in R+$. So $f(z)+n=f(z+n)$ for all $n \in N$, $z \in R+$............(11) Collary of (11) is $f(n)=n$ for all $n \in N$. In (**), take $a=b \in N$ and use (11), then $[f(c)+1][2a]=2[f(ac)+a]$. So $af(c)=f(ac)$ for all $a \in N, c \in R+$............(12) Collary of (12) is $f(q)=q$ for all $q \in Q+$, by taking $c=p/q, a=q$. In (**) again, choose $a=b$, by (11) we have $2f(a)=f(2)f(a)=f(2a)$. So $2f(ac+a)=[f(c)+1]f(2a)=2[f(c)+1]f(a)$ => $f(ac+a)=f(a)f(c)+f(a)$. Now for any $d>0$, take $e=d/x$, then $f(x+d)=f(x+xe)=f(x)f(e)+f(x)>f(x)$. So $f(x+d)>f(x)$ for all $d>0$ => f is strictly increasing.........(13) Now suppose $r \in R+$ and $f(r)<r$, we can take some $q\in Q+$ with $f(r)<q<r$. So $f(r)<q=f(q)$ while $q<r$, false by (13). For $f(r)>r$, this is handled similarly. So it must be $f(r)=r$ for all $r\in R+$ Check: $(z+1)(x+y)=(xz+y)+(yz+x)$, good. Only Solution: $f(x)=x$ for all $x\in R+$

Just for completeness: Here is the solution I posted in a similar thread a month ago...before it was deleted

Lengthy solution, core idea is: prove that $f$ is increasing and has infinitely many fixed points.

ugh nonnegatives to nonnegatives is so ez i cant read

navi_09220114. Why do you know that the function f is surjective on ([Z+1]/2×f(2), infinity)?

Opps sorry, that was a typo, i meant $(\frac{1}{2}f(2),+\infty)$. Its because for any $x$ in that interval, we can find some $z>0$ with $f(f(z)+1)=\frac{(z+1)}{2} \cdot f(2)=x$, so $x \in Im(f)$ for these $x$.

Gurtonn wrote: if $x=y$ we get $2f(xf(z)+x)=(z+1)f(2x) \Rightarrow f(xf(z)+x)=\frac{z+1}{2} \cdot f(2x)$ therefore if $a > \frac{1}{2}$ and $b\in Im(f)$ then $ab \in Im(f)$, Proving that $f$ is surjective. let $a$ be a number s.t $f(a) = 1$, then $(a+1)f(x+y) = f(xf(a)+y)+f(yf(a)+x)= 2f(x+y) \Rightarrow a=1$, so $f(1) = 1$ then $z+1 = (z+1)f(\frac{1}{2} + \frac{1}{2})=2f(\frac{f(z)+1}{2}) \Rightarrow f(\frac{f(z)+1}{2}) = \frac{z+1}{2}$ substituting $z=\frac{z+1}{2}$ we get $f(\frac{z+3}{4})=\frac{z+3}{4}$, so for $x>\frac{3}{4}, f(x)=x$. let $z$ be any number, and $x,y$ sufficiently big numbers so $x+y,xf(z)+y,yf(z)+x > \frac{3}{4}$, then: $(z+1)(x+y)=(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x)=xf(z)+y+yf(z)+x =(f(z)+1)(x+y) \Rightarrow f(z) = z$ I'm afraid the fifth line is wrong. How can you get $f(\frac{z+3}{4})=\frac{z+3}{4}$ ?

Hope I'm correct If there is a sequence $a_n>0$ satisfies: 1.$\lim_{n\rightarrow\infty}a_n=0$ 2.$\lim_{n\rightarrow\infty}f(a_n)=\infty$ We can WLOG assume all $a_n<1$ by forgetting first few terms, plug in $y=1-x,z=a_n$ in the original equation we can get: \[f(xf(a_n)+(1-x))<(a_n+1)f(1)<2f(1)\]Which tells us $f((1,f(a_n)))$ bounded by $2f(1)$, take $n$ goes to infinite and get $f((1,\infty))$ is bounded by $2f(1)$, but if we take $x>1,y>1$ and $z$ very large in the original equation, RHS is smaller than $4f(1)$, but LHS can be arbitrary large, hence we get a contradiction! Which tells us such sequence $a_n$ don't exist, hence $\limsup_{x\rightarrow 0^+}f(x)$ exist. Assume $\limsup_{x\rightarrow 0^+}f(x)=k$, then we can find a sequence $x_n>0$ such that: 1.$\lim_{n\rightarrow\infty}x_n=0$ 2.$\lim_{n\rightarrow\infty}f(x_n)=k$ Plug in $x=y=\frac{x_n}{2}$ and take $n\rightarrow\infty$, we'll get \[(z+1)k=\lim_{n\rightarrow\infty}(z+1)f(x_n)=\lim_{n\rightarrow\infty}2f(x_nf(z)+x_n)\leq 2k\]So $k=0$, and hence $\lim_{z\rightarrow 0^+}f(z)=0$. $f(x+y)\geq \frac{f(xf(z)+y)}{1+z} $ If exist $a>b, f(a)<f(b)$ solve $x+y=a,xf(z)+y=b$ $x=\frac{a-b}{1-f(z)}$ $y=\frac{b-af(z)}{1-f(z)}$ and by $\lim_{z\rightarrow 0}f(z)=0$ we can choose $z$ very small and $f(z)$ very small and get a contradiction! so $f$ is an increasing function. take $z\rightarrow 0$ and let $g(x)=\lim_{h\rightarrow 0^+}f(x+h)$ we have $f(x+y)=g(x)+g(y)$ take $y$ as $y+\varepsilon$ and $\varepsilon\rightarrow 0^+$ $g(x+y)=g(x)+g(y), g\geq 0$ so $g(x)=cx$ and then $f(x)=cx$ and get $c=1$

shinichiman wrote: Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $$(z + 1)f(x + y) = f(xf(z) + y) + f(yf(z) + x),$$for all positive real numbers $x, y, z$. Plug in $x=y$ to find that \[ (z+1)f(2x)=2f\big(xf(z)+x\big).\qquad (1). \]Clearly, $f$ is injective, and running $z$ over $\mathbb{R}^+$ in $(1)$, we see that $f$ attains every value $>\frac{f(2x)}{2}$, $\forall x$, from which it can easily be deduced that $\inf f=0$, so $f$ is surjective. If $f(z)=1$, then $(1)$ gives $z=1$; since $f$ is bijective, $f(1)=1$ follows. Putting $x=\frac12$ in $(1)$ yields $f\left(\frac{f(z)+1}{2}\right)=\frac{z+1}{2}$. Plugging $z=\frac{f(4x-3)+1}{2}$ into this and rearranging gives $f(4x-3)=4f(x)-3$ whenever $x>\frac34$. Hence, \[ \boxed{f(x)>\frac34\text{ if }x>\frac34.} \] Moreover, $(1)$ shows that $\frac{f(cX)}{f(X)}$ is constant whenever $c=\frac{f(z)+1}{2}$. Due to surjectivity, $\frac{f(cX)}{f(X)}$ is actually constant for all $c>1$ (and thus for all $c<1$, too). Therefore, using $f(1)=1$, we get $\boxed{f(a)f(b)=f(ab)}$. The two boxed statements imply that $h(x)=\ln f(e^x)$ must be linear by Cauchy's equation, and so $f(x)=x^\alpha$ for some $\alpha\in\mathbb{R}$. Finally, from $(1)$ for $x=1$, we get the identity $(X+1)2^\alpha=2(X^\alpha+1)^\alpha$. The finishing touches: injectivity and $f(x)>\frac34$ if $x>\frac34$ imply $0<\alpha\le 1$; then taking the limit of both sides as $X\to 0$ shows $\alpha=1$. Therefore, $f(x)=x$ for all $x$, done. $\blacksquare$

Suppose $f(z_1) = f(z_2)$. Then throw in $z_1$ and $z_2$. The RHS remains invariant, and the LHS changes, so this is a contradiction. Thus $f$ is injective. Take $z = 1, x = y$ to obtain $f(2x) = f(xf(z)+x)$. If $f(z) \neq 1$, then $f$ is not injective, so $f(1) = 1$. Now if we fix $x = y$ and vary $z$, we see that if $f(2x) = k$, then we can attain any value in $(k/2, \inf)$ by varying $z$, so $f$ is surjective. Now let $a = xf(z) + y, b = yf(z) + x$. Then the original equation rearranges to $f(a) + f(b) = (z + 1) f( (a + b) / (1 + f(z)))$. In particular, for $z = 1$ we get Jensen's FE. Take $g(x) = f(x + 1) - 1$ with domain and range $(-1, \inf)$. Then we know $g(a) + g(b) = g(a + b)$. Observe that if $g(x_0) < 0$ for some $x_0 > 0$, we would die because $\lim_{k \rightarrow \inf} g(kx_0) \rightarrow -\inf$ by additive, which contradicts the fact that the range is $(-c, \inf)$. Hence, $g(x) \ge 0$ for $x > 0$. Now if we take any $b > 0$, we get $g(a) + g(b) = g(a + b)$, but $g(b) > 0$ by injectivity so $g(a) < g(a + b)$. This implies $g$ is increasing. Since $g$ is increasing and additive, it follows that it is linear by using the density of the rationals in the reals. So $f$ is also linear. I'm too lazy to find the answer.

Here is a relatively clean solution. The only function is $f(x)=x$, which clearly is a solution to the equation. First, note that $f$ is injective. Indeed, setting $z=z_1$ and $z=z_2$ and supposing that $f(z_1)=f(z_2)=c$ gives that $z_1=z_2=\frac{f(cx+y)+f(cy+x)}{f(x+y)}-1$. Next, setting $x=y=z=1$ gives $2f(2)=2f(f(1)+1)$, so by injectivity $f(1)=1$. Now, setting $x=y$ gives $(z+1)f(2x)=2f(x(f(z)+1))$, so for any $u$ in the range of $f$, $\frac{u(z+1)}2$ is also in the range of $f$ for any $z$. Surjectivity easily follows. Suppose that $a$ and $b$ are distinct positive reals. Setting $x=\frac{af(z)-b}{f(z)^2-1}$ and $y=\frac{bf(z)-a}{f(z)^2-1}$ with $1\ne f(z)\in\left(\frac ab,\frac ba\right)$ gives that $(z+1)f\left(\frac{a+b}{f(z)+1}\right)=f(a)+f(b)$. Taking $z$ with $|f(z)-1|<\epsilon$ for a sufficiently small $\epsilon$ gives that $f(a)+f(b)=f(c)+f(d)$ for any $a\ne b,c\ne d$ with $a+b=c+d$. In particular, we have for $x$ and $y$ distinct that $f(4x)+f(x+3y)=f(3x+y)+f(2x+2y)$ and $f(3x+y)+f(4y)=f(2x+2y)+f(x+3y)$. Adding gives $f(4x)+f(4y)=2f(2x+2y)$, so $f(x)+f(y)=2f\left(\frac{x+y}2\right)$ for all $x,y\in\mathbb R^+$. Thus, $f(a)+f(b)=f(c)+f(d)$ for any $a+b=c+d$, so for any rationals $q$ and $r$ with $qx+ry>0$ and $q+r=1$ we have that $qf(x)+rf(y)=f(qx+ry)$. In other words, given two distinct numbers $a,b,c$, $(c,f(c))$ will lie on the line through $(a,f(a))$ and $(b,f(b))$ provided that $c$ is a rationally weighted average of $a$ and $b$. If such a line has a positive root, there is some interval in which it is negative, so we can rationally weight $a$ and $b$ into that interval resulting as $\mathbb Q$ is dense in $\mathbb R$ resulting in a negative output of the function, a contradiction. Hence, all such lines must have nonpositive roots. In particular, this means that $f$ is increasing and that $x>1\implies f(x)\le x$ and $x<1\implies f(x)\ge x$. Finally, take $x+y=1$ and $z>1\implies 1<f(z)\le z$. Then, $z+1=(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x)\le xf(z)+y+yf(z)+x=f(z)+1$ as $xf(z)+y,yf(z)+x>1$, so $f(z)\ge z$. This means that $f(z)=z$ for $z>1$. Setting $z>1$ in $(z+1)f(2x)=2f(x(f(z)+1))$ gives that $(z+1)f(2x)=2f(x(z+1))$, so $f(x)=x\implies f\left(\frac{2x}{z+1}\right)=\frac{2x}{z+1}$ for all $z>1,x$, implying that $f(x)=x$ for all $x\in\mathbb R$. @below thanks for pointing out the error

ABCDE wrote: $f\left(\frac2{f(z)+1}\right)=\frac2{z+1}$. This means that $f$ is surjective. I can't understand how this covers all reals, as $z>0$ gives only reals in $(0,2)$.

As usual, let $P(x,y,z)$ denote $(z+1)f(x + y) = f(xf(z) + y) + f(yf(z) + x)$ for all $x,y,z\in \mathbb{R}^+$. First, it's not hard to see that $f$ is surjective on $\mathbb{R}^+$. For all $t,x,y,z\in \mathbb{R}^+$, $P(t+x,y,z)$ and $P(x,t+y,z)$ gives us $$Q(t,x,y,z): f(tf(z)+xf(z)+y)+f(yf(z)+x)=(z+1)f(t+x+y)=f(xf(z)+y)+f(tf(z)+yf(z)+x).$$Now, for any $A,B,C\in \mathbb{R}^+$, since $f$ is surjective, there exists $r\in \mathbb{R}^+$ that $f(r)>\max \{ \frac{A}{B},1,\frac{B}{A}\}$. Let $p=\frac{Af(r)-B}{f(r)^2-1}$ and $q=\frac{Bf(r)-A}{f(r)^2-1}$, both are positive real numbers. There also exists $s\in \mathbb{R}^+$ that $s=\frac{C}{f(r)}$. It's not hard to verify that $A=pf(r)+q,B=qf(r)+p,$ and $C=sf(r)$. Then, $Q(s,p,q,r)$ gives us $f(C+A)-f(A)=f(C+B)-f(B)$. This means that, for all $c\in \mathbb{R}^+$, there exists $g(c) \in \mathbb{R}$ that $f(c+x)-f(x)=g(c)$ for all $x\in \mathbb{R}^+$. Since, for all $c,d\in \mathbb{R}^+$, $g(c+d)=f(c+d+1)-f(1)=\left( f(c+d+1)-f(d+1)\right) +\left( f(d+1)-f(1)\right)$. We get $g(c+d)=g(c)+g(d)$ for all $c,d\in \mathbb{R}^+$. $P(x,y+c,z)$ gives us $$(z+1)g(c)+(z+1)f(x+y)=g(c)+f(xf(z)+y)+f(cf(z)+yf(z)+x)$$for all $x,y,z,c\in \mathbb{R}^+$. So, combining with $P(x,y,z)$, we get $$zg(c)+f(yf(z)+x)=f(cf(z)+yf(z)+x)\implies g(cf(z))=zg(c)$$for all $x,y,z,c\in \mathbb{R}^+$. Hence, $f(xf(z)+y)=f(y)+g(xf(z))=f(y)+zg(x)$ and $f(yf(z)+x)=f(x)+g(yg(z))=f(x)+zg(y)$ for all $x,y,z\in \mathbb{R}^+$. So, $P(x,y,z)$ becomes $$zf(x+y)+f(x+y)=z\left( g(x)+g(y)\right) +\left( f(x)+f(y)\right)$$for all $x,y,z\in \mathbb{R}^+$. This easily gives $f(x+y)=g(x)+g(y)=g(x+y)$ and $f(x+y)=f(x)+f(y)$ for all $x,y\in \mathbb{R}^+$. Back to $g(cg(z))=g(cf(z))=zg(c)$ for all $z,c\in \mathbb{R}^+$. Note that we also have verified that $g(z)=f(z)\in \mathbb{R}^+$ for all $z\in \mathbb{R}^+$. We get $g(g(c)g(z))=zg(g(c))$, and similarly, $g(g(c)g(z))=cg(g(z))$ for all $z,c\in \mathbb{R}^+$. Hence, there exists $\ell \in \mathbb{R}^+$ that $g(g(x))=x\ell$ for all $x\in \mathbb{R}^+$. Hence, $g(\ell cz)=g(cg(g(z)))=g(z)g(c)$ for all $z,c\in \mathbb{R}^+$. In other words, $g(ab/\ell ) =g(a/\ell )g(b/\ell )$ for all $a,b\in \mathbb{R}^+$. Let $h:\mathbb{R}^+\rightarrow \mathbb{R}^+$ be a function defined by $h(x)=g(x/\ell )$ for all $x\in \mathbb{R}^+$. Easy to verify that $h$ is both multiplicative and additive, so $h$ must be identity function. So, $f(x)=g(x)=x\ell$ for all $x\in \mathbb{R}^+$. Plugging in the initial problem gives $f(x)=x$ for all $x\in \mathbb{R}^+$, which, hence, is the only solution.

Nice problem!! My solution: Let $P(x,y,z)$ denote the given assertion. LEMMA-1: $f$ is injective.

LEMMA-2: $f(1) = 1$

LEMMA-3: $f$ is surjective.

LEMMA-4: $f\left(\frac{1}{x} \right) = \frac{1}{f(x)}$

LEMMA-5: $f$ is multiplicative.

LEMMA-6: $f$ is involutive, i.e. $f(f(x)) = x \text{ } \forall x \in \mathbb{R^+}$

LEMMA-7: $f(2) = 2$

LEMMA-8: $f$ is additive.

LEMMA-9: $f(x) = x$

Hence, done

We claim that the only solution is $f(x) \equiv x$ for all $x \in \mathbb{R}^{+}.$ It is trivial to verify that this solution indeed satisfies the equation. We'll now show a series of lemmas. $\mathbf{Lemma}$ $\mathbf{1.}$ $f(z)$ can take on arbitrarily small values. $\mathbf{Proof.}$ By plugging in $z = \frac13,$ we know that: $$f(xf(\frac13)+y) + f(yf(\frac13)+x) = \frac43 f(x+y).$$Therefore, for any $z \in \mathbb{R}^{+},$ by letting $x + y = z$, we can obtain another value $z' \in \mathbb{R}^{+}$ such that $f(z') \leq \frac23 f(z).$ From this, it is clear that we can get $f(z)$ to tend arbitrarily close to $0$, as desired. $\blacksquare$ $\mathbf{Lemma}$ $\mathbf{2.}$ $f(x) + f(y) = f(t) +f(z)$ when $x+y = t+z$. $\mathbf{Proof.}$ Consider fixing some $z$ for which $f(z)$ is arbitrarily small. Then, observe that when we fix $s$, we have that $f(a) + f(b)$ is fixed whenever $sf(z) \leq a, b\leq s$ and $a+b = sf(z) + s$. Since $f(z)$ is tiny, we can make $s$ such that $sf(z) + s = x+y = t+z$ and $sf(z) \leq x, y, t, z \leq s$. Therefore, we are done. $\blacksquare$ As a direct corollary of the above, observe that Jensen's Functional Equation $f(\frac{x+y}{2}) = \frac{f(x) + f(y)}{2}$ now holds. Therefore, this furthermore implies that $f$ is nondecreasing by Jensen's Functional Equation. Since $f$ is $\mathbb{R}^{+} \rightarrow \mathbb{R}^{+},$ it's easy to see that $f$ is linear. To see this, observe that for any two values $a$ and $b$, then line through $(a, f(a))$ and $(b, f(b))$ has a dense subset which are points on $f.$ Therefore, for any point $(x, f(x))$ which isn't on the line, we can clearly use the Jensen condition to get a negative value of $f$ (take a point $(y, f(y))$ on the line for which $y$ is arbitrarily close to $y$ on whichever side, depending on whether $(x, f(x))$ is over/under the line).Therefore, we obtain that $f(x) \equiv ax + b$. Plugging this into the equation, it's easy to see that only $f(x) \equiv x$ satisfies the desired conditions. $\square$

Partial solution: I will prove that $f(x)f(y)=f(xy)$ and $f(x)+f(y)=2f(\frac{x+y}{2})$. Here $P(x,y,z)$ is the assertion that $(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x)$ Proof: Claim 1: $f(1)=1$. Suppose $f(1)=k \neq 1$. Then $P(\frac{x}{2},\frac{x}{2},1)$ gives $f(x)=f(x(\frac{k+1}{2})$. Now put $P(\frac{x}{2},\frac{x}{2},\frac{k+1}{2})$ to get a contradiction( basically getting that two unequal numbers are equal). Hence, $f(1)=1$. Claim 2:$\frac{z+1}{2}=f(\frac{f(z)+1}{2})$. Proof:$P(\frac{1}{2},\frac{1}{2},z)$ with Claim 1 gives the result. Claim 3:$f(x)$ is multiplicative and surjective. Proof: Because of claim 2,$f(x)$ can attain any value more than $0.5$. Now, put $P(\frac{x}{2},\frac{x}{2},z)$ to get $\frac{z+1}{2}f(x)=f(x\frac{f(z)+1}{2})$ or $f(\frac{f(z)+1}{2})f(x)=f(x\frac{f(z)+1}{2})$. As $f(x)$ can obtain any value more than $0.5$, we can see that $f(x)f(y)=f(xy)$ if one of them is higher than $0.75$. Now consider this: Let $x,y \leq 0.75$. Then, obviously $f(y)f(\frac{1}{y})=1$ as $f(\frac{1}{y})$ is higher than $0.75$. Now, $f(xy)f(\frac{1}{y})=f(x)$ which from previous result is same as $f(x)f(y)=f(xy)$ even in the case when both are less than $0.75$. Hence, $f(x)f(y)=f(xy)$ for all reals. Now use this fact combined with that if $k>0.5$, then there exists real $x$ with $f(x)=k$ to prove that $f(x)$ is surjective. Claim 4:$f(x)+f(y)=2f(\frac{x+y}{2})$ Proof: $P(x,y,z)$ with Claim 3 gives that $f(xf(z)+y)+f(yf(z)+x)=2f((f(z)+1)\frac{x+y}{2})$. Now use $f(x)$ being surjective to prove that for any given pair of positive reals $b,c$ one can find positive reals $x,y,z$ with $xf(z)+y=b,yf(z)+x=c$ which gives $f(b)+f(c)=2f(\frac{b+c}{2})$. I am pretty sure that after proving Jensen's FE condition and multiplicity with boundedness($f(x)$ is always a positive real), proving that $f(x)=x$ should be trivial but I can't put it all together so this was all the progress I made.

Continuing my proof: From Jensen's FE, I claim that $f(x)$ is monotonically increasing. Suppose this is not true. One can find $a<b$ such that $f(a)>f(b)$. One can see that $f(nb-(n-1)a)=nf(b)-(n-1)f(a)$. The LHS is always positive but the RHS eventually becomes negative, contradiction. Hence $f(x)$ is monotonically increasing. From Jensen's FE and multiplicity one can obtain that $f(\frac{k}{2^n})=\frac{k}{2^n}$ where $k,n \in N$. Using that $f(x)$ monotonically increasing, one can bound the value of $f(x)$ for any real. One can show that $|f(x)-x|$ is smaller than any given real for any $x$ hence $f(x)=x$ for all $x$.

shinichiman wrote: Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $$(z + 1)f(x + y) = f(xf(z) + y) + f(yf(z) + x),$$for all positive real numbers $x, y, z$. Fajar Yuliawan, Indonesia UwU nice problem. We claim that all solutions to the given functional equation are of the form $\boxed{f(x) = x}$ for all positive reals $x$. Let $P(x, y)$ denote the statement of the problem. Step 1 : $f(1) = 1$ and $f$ is injective. Of course observe that if $f(a) = f(b)$ for some positive reals $a$ and $b$, then comparing $P(k, k, a)$ and $P(k, k, b)$ yields that $a + 1 = b +1$ or $a = b$ proving that $f$ is injective. Now see that $P(1, 1, 1)$ implies that $f(1) = 1$. Step 2 : $f$ is bijective, additive and multiplicative which essentially implies our desired result We see that using $P(z, z, 1)$, $f$ is bijective as we've already proved that it is injective. Now, see that if $f(y) = t-1$ for some real $t > 1$, then $f(tx) = \frac{f(2)f(x)f(y+1)}{2}$ and we see that $f(2x) = f(2)f(x)$ holds true (using basic assertions) which means that in fact $f(tx) = f(x)f(t)$ for $t > 1$ which isn't difficult to see that it is true for $t > 0$ too and so $f$ is multiplicative. Now, observe that using multiplicativity and results of $P(x, x, z)$ and $P(x, x, z^{-1})$, we yield that $f(f(x)) = x$. Now, $P(1, 1, z) \implies f(f(z) + 1) = \frac{(z+1)f(2)}{2}$ and now see that $f(2)f(z) + f(2) = 2f(z+1)$. If $f(2) = 2a$, then see that $f(4) = 4a^2 = 2a^3 + a^2 + a$ using basic assertions and if $a = \frac{1}{2}$, it would contradict bijectivity and $f(2) > 0$ so $a = 1$ and $f(2) = 2$ which immediately proves $f(z) + 1 = f(z +1)$. Not hard to see that $f(x+y) = f(x)f(1 + y/x) = f(x) + f(y)$ using all above assertions which means that $f$ is additive. But $f = multiplicative$, $f = additive$ $\Longrightarrow f(x) = x$ for all positive reals $x$ which clearly satisfies the conditions of the given problem and we're done.

The solution is $f(x)=x$ which obviously works. We now show that it is the only one. Label the equation $$(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x)\hspace{20pt}(1)$$ Lemma. If $a>b,c>1$ are positive reals then there exists positive reals $x,y$ such that $$\begin{cases}xc+y&=a\\ yc+x&=b\end{cases}$$if and only if $c>\frac{a}{b}$. Proof. Solve the equation to obtain $$x=\frac{ac-b}{c^2-1}, y=\frac{bc-a}{c^2-1}$$then $x,y>0$ if and only if $c>\frac{a}{b}$ Claim 1. $f$ is unbounded and injective. Proof. Suppose on the contrary that $f$ is bounded, then fixing $x,y$ and let $z$ tends to infinity, then $L.H.S.$ is unbounded while $R.H.S.$ is bounded, contradiction. Now, suppose $f(a)=f(b)$, substituting $z=a$ and $z=b$ into $(1)$ we have $$(a+1)f(x+y)=(b+1)f(x+y)$$hence $a=b$. $\blacksquare$ Claim 2. If $a+b=c+d$ and $a>b,c>d$ then $$f(a)+f(b)=f(c)+f(d)$$Proof. Indeed, pick some $z$ such that $f(z)>\max\{\frac{a}{b},\frac{c}{d}\}$, by the lemma we can select $x_1,y_1,x_2,y_2$ such that $x_1f(z)+y_1=a$, $y_1f(z)+x_1=b$, $x_2f(z)+y_2=c$, $y_2f(z)+x_2=d$, then by $(1)$ we have $$f(a)+f(b)=(z+1)f(x_1+y_1)=(z+1)f(\frac{a+b}{f(z)+1})=(z+1)f(\frac{c+d}{f(z)+1})=(z+1)f(x_2+y_2)=f(c)+f(d)$$as desired. $\blacksquare$ As a corollary, $$(z+1)f(x+y)=f((x+y)f(z))+f(x+y)$$and so $$zf(a)=f(af(z))\hspace{20pt}(2)$$In particular, substituting $z=1$ in $(2)$ we have $f(1)=1$ by injectivity. Substituting $x=y$ in $(1)$ and using $(2)$, $$2xf(z+1)=f(2)x(f(z)+1) \hspace{20pt}(3)$$Hence substituting $z=2$ in $(3)$ we have $f(2)^2+f(2)=2f(3)=f(2)+f(4)$, hence $f(4)=f(2)^2$, solving \begin{align*} 1+f(3)&=2f(2)\\ f(2)+f(2)^2=2f(3) \end{align*}We have $f(2)=1$ or $2$, hence $f(2)=2$ by injectivity, from $(3)$ we have $f(z+1)=f(z)+1$, so $f(x)=x$ for all positive integer $x$, moreover from $(2)$ we have $$f(x)+f(y)=2f(\frac{x+y}{2})=f(x+y)$$Hence by Cauchy equation we can conclude that $f(x)=x$.

Solved with Rama1728 (we spend 1 hour on this, and it was worth it!) Let $P(x,y)$ the assertion of the given F.E. Claim 1: $f$ is bijective Proof: Let $a,b$ such that $f(a)=f(b)$, thus by $P(1,1,a)-P(1,1,b)$ $$f(2)(b+1)=f(2)(a+1) \implies a=b \implies f \; \text{injective}$$$P \left(\frac{1}{f(x)+1},\frac{1}{f(x)+1},x \right)$ $$f \left(\frac{2}{f(x)+1} \right)=\frac{2}{x+1} \implies f \; \text{surjective on} \; [0,2]$$$P \left(\frac{1}{2},\frac{1}{2},x \right)$ $$f \left(\frac{f(x)+1}{2} \right)=\frac{x+1}{2} \implies f \; \text{surjective on} \; \left[\frac{1}{2}, \infty \right]$$Thus $f$ is bijective as desired. Claim 2: $f$ is multiplicative. Proof: Let $c=\frac{f(y)+1}{2}$, by $P \left(\frac{x}{2},\frac{x}{2},y \right)$ $$f(c)f(x)=f(cx) \implies f \; \text{multiplative with} \; c>\frac{1}{2}$$Now note that $f \left(\frac{1}{c} \right)=\frac{1}{f(c)}$ thus by $P \left(\frac{x}{f(y)+1},\frac{x}{f(y)+1},y \right)$ $$f \left(\frac{x}{c} \right)=f(x)f \left(\frac{1}{c} \right) \implies f \; \text{multiplicative with} \; \frac{1}{c}<2$$Meaning that $f$ is multiplicative for all positive reals. Claim 3: $f$ is continuos Proof: By Claim 1 it suffices to show that $\lim_{d \to d_0} f \left( \frac{f(d)+1}{2} \right)=f \left( \frac{f(d_0)+1}{2} \right)$ but we have: $$\lim_{d \to d_0} f \left(\frac{f(d)+1}{2} \right)=\lim_{d \to d_0} \frac{d+1}{2}=\frac{d_0+1}{2}=f \left(\frac{f(d_0)+1}{2} \right)$$Thus $f$ is continuos as desired. Main solution: By Claim 2 and Claim 3 we have that $f(x)=x^k$ and now let $f(2)=2s$, we have: $$f(x+1)=s(f(x)+1) \implies f(x+2)=s(f(x+1)+1)=s^2f(x)+s^2+s \overset{x=2}{\implies} 4s^2=f(4)=s(2s^2+s+1) \implies s=1$$The last step bc $s=\frac{1}{2}$ is also a solution to that but by Claim 1 we would have $f(2)=f(1)$ so $2=1$ which makes nonsense. Since $f(2)=2$ we have that $2^k=f(2)=2$ thus $k=1$. Meaning that the unique sol is $f(x)=x$ and we are done

MathLuis wrote: Solved with Rama1728 (we spend 1 hour on this, and it was worth it!) Let $P(x,y)$ the assertion of the given F.E. Claim 1: $f$ is bijective Proof: Let $a,b$ such that $f(a)=f(b)$, thus by $P(1,1,a)-P(1,1,b)$ $$f(2)(b+1)=f(2)(a+1) \implies a=b \implies f \; \text{injective}$$$P \left(\frac{1}{f(x)+1},\frac{1}{f(x)+1},x \right)$ $$f \left(\frac{2}{f(x)+1} \right)=\frac{2}{f(x)+1} \implies f \; \text{surjective on} \; [0,2]$$$P \left(\frac{1}{2},\frac{1}{2},x \right)$ $$f \left(\frac{f(x)+1}{2} \right)=\frac{x+1}{2} \implies f \; \text{surjective on} \; \left[\frac{1}{2}, \infty \right]$$Thus $f$ is bijective as desired. Claim 2: $f$ is multiplicative. Proof: Let $c=\frac{f(y)+1}{2}$, by $P \left(\frac{x}{2},\frac{x}{2},y \right)$ $$f(c)f(x)=f(cx) \implies f \; \text{multiplative with} \; c>\frac{1}{2}$$Now note that $f \left(\frac{1}{c} \right)=\frac{1}{f(c)}$ thus by $P \left(\frac{x}{f(y)+1},\frac{x}{f(y)+1},y \right)$ $$f \left(\frac{x}{c} \right)=f(x)f \left(\frac{1}{c} \right) \implies f \; \text{multiplicative with} \; \frac{1}{c}<2$$Meaning that $f$ is multiplicative for all positive reals. Claim 3: $f$ is continuos Proof: By Claim 1 it suffices to show that $\lim_{d \to d_0} f \left( \frac{f(d)+1}{2} \right)=f \left( \frac{f(d_0)+1}{2} \right)$ but we have: $$\lim_{d \to d_0} f \left(\frac{f(d)+1}{2} \right)=\lim_{d \to d_0} \frac{d+1}{2}=\frac{d_0+1}{2}=f \left(\frac{f(d_0)+1}{2} \right)$$Thus $f$ is continuos as desired. Main solution: By Claim 2 and Claim 3 we have that $f(x)=x^k$ and now let $f(2)=2s$, we have: $$f(x+1)=s(f(x)+1) \implies f(x+2)=s(f(x+1)+1)=s^2f(x)+s^2+s \overset{x=2}{\implies} 4s^2=f(4)=s(2s^2+s+1) \implies s=1$$The last step bc $s=\frac{1}{2}$ is also a solution to that but by Claim 1 we would have $f(2)=f(1)$ so $2=1$ which makes nonsense. Since $f(2)=2$ we have that $2^k=f(2)=2$ thus $k=1$. Meaning that the unique sol is $f(x)=x$ and we are done A small change to be made, the surjectivity part is incorrect, instead we can do \(P(x,x,y)\) to get that \(x\in\text{Im}(f)\) implies \(y\in\text{Im})(f)\) for all \(2y>x\) so we get surjectivity. And instead of continuity, we prove monotonicity, because something does not seem rigorous in the continuity part. How we prove monotonicity is by doing \(P(x/2,x/2,y)\) which gives us \[f(\frac{x(f(y)+1)}{2})=\frac{(y+1)f(x)}{2}\]and we compare the domains on both sides. We get that \[f\left((\frac{x}{2},\infty\right)=(f(\frac{x}{2},\infty)\]which proves our desired claim. From this we easily get \(f(x)=x^c\) because \(f\) is multiplicative...

MathLuis wrote: Solved with Rama1728 (we spend 1 hour on this, and it was worth it!) Let $P(x,y)$ the assertion of the given F.E. Claim 1: $f$ is bijective Proof: Let $a,b$ such that $f(a)=f(b)$, thus by $P(1,1,a)-P(1,1,b)$ $$f(2)(b+1)=f(2)(a+1) \implies a=b \implies f \; \text{injective}$$$P \left(\frac{1}{f(x)+1},\frac{1}{f(x)+1},x \right)$ $$f \left(\frac{2}{f(x)+1} \right)=\frac{2}{x+1} \implies f \; \text{surjective on} \; [0,2]$$$P \left(\frac{1}{2},\frac{1}{2},x \right)$ $$f \left(\frac{f(x)+1}{2} \right)=\frac{x+1}{2} \implies f \; \text{surjective on} \; \left[\frac{1}{2}, \infty \right]$$Thus $f$ is bijective as desired. Claim 2: $f$ is multiplicative. Proof: Let $c=\frac{f(y)+1}{2}$, by $P \left(\frac{x}{2},\frac{x}{2},y \right)$ $$f(c)f(x)=f(cx) \implies f \; \text{multiplative with} \; c>\frac{1}{2}$$Now note that $f \left(\frac{1}{c} \right)=\frac{1}{f(c)}$ thus by $P \left(\frac{x}{f(y)+1},\frac{x}{f(y)+1},y \right)$ $$f \left(\frac{x}{c} \right)=f(x)f \left(\frac{1}{c} \right) \implies f \; \text{multiplicative with} \; \frac{1}{c}<2$$Meaning that $f$ is multiplicative for all positive reals. Claim 3: $f$ is continuos Proof: By Claim 1 it suffices to show that $\lim_{d \to d_0} f \left( \frac{f(d)+1}{2} \right)=f \left( \frac{f(d_0)+1}{2} \right)$ but we have: $$\lim_{d \to d_0} f \left(\frac{f(d)+1}{2} \right)=\lim_{d \to d_0} \frac{d+1}{2}=\frac{d_0+1}{2}=f \left(\frac{f(d_0)+1}{2} \right)$$Thus $f$ is continuos as desired. Main solution: By Claim 2 and Claim 3 we have that $f(x)=x^k$ and now let $f(2)=2s$, we have: $$f(x+1)=s(f(x)+1) \implies f(x+2)=s(f(x+1)+1)=s^2f(x)+s^2+s \overset{x=2}{\implies} 4s^2=f(4)=s(2s^2+s+1) \implies s=1$$The last step bc $s=\frac{1}{2}$ is also a solution to that but by Claim 1 we would have $f(2)=f(1)$ so $2=1$ which makes nonsense. Since $f(2)=2$ we have that $2^k=f(2)=2$ thus $k=1$. Meaning that the unique sol is $f(x)=x$ and we are done Wrong solution .You can say $f(\frac{f(x)+1}{2})$ is continous not $f(x)$.

$f(x)=x$ is the only solution to the FE which clearly works. Now we show this is the only solution. Let $P(x,y,z)$ denote the given FE.

Now back to our main solution. Observe that compairing $P(xf(z),yf(z),1/z)$ and $P(x,y,z)$ gives us $$f(xf(z)+yf(z))=zf(x+y).........(5)$$ Now fix $x+y=1$ and we get $f(f(z))=z$ for all real number $z$. So actually $f$ is surjective function. Exchanging $x,y$ with $x/2,x/2$ and $z$ with $f(z)$ gives us $f(z)f(x)=f(xz)$. So $f$ has multiplicative property. Using the multiplicative property from $P(x,x,z)$ we get that $z+1=f(f(z)+1)$. For any positive real number $a,b$ we can fix $f(z)$ and choose $x,y$ such that $xf(z)+y=a$ and $yf(z)+x=b$. Observe, $$f(a+b)=(z+1)f(x+y)=f(f(z)+1)f(x+y)=f(xf(z)+y+yf(z)+x)=f(a)+f(b)$$ That means $f$ has additive property too. Since $f$ is additive and multiplicative, $f(x)=x$ is the only solution.

So we're done.

Denote the assertion by $P(x,y,z).$ Claim: $u+v=s+t \implies f(u)+f(v)=f(s)+f(t).$ Proof. We have surjectivity from induction on $P(x,x,z).$ So it is trivial that we may find the numbers \begin{align*} u=af(x)+b \qquad v=bf(x)+a \\ s=cf(x)+d \qquad t=df(x)+c\end{align*}Now $P(a,b,x)$ and $P(c,d,x)$ yield the claim. $\blacksquare$ In particular $f$ satisfies Jensen's Functional Equation, which we can solve (as we are in $\mathbb{R}^+$) to get $f(x)=ax+b.$ Checking gives $b=0, a=1.$ We see that the identity satisfies.

Claim $1 : f$ is injective. Proof $:$ if $f(z) = f(z')$ then $P(x,y,z) - P(x,y,z')$ implies that $(z-z')f(x+y) = 0$ and $f(x+y) \neq 0$ so $z = z'$ Claim $2 : f(1) = 1$. Proof $:$ Assume $f(1) = c$ then $P(1,1,1) : 2f(2) = 2f(c+1) \implies f(2) = f(c+1) \implies c = 1$ Claim $3 : f$ is bijective for all $x > \frac{1}{2}$. Proof $:$ $P(\frac{1}{2},\frac{1}{2},z) : \frac{z+1}{2}f(1) = f(f(z)+1) \implies f(f(z)+1) = \frac{z+1}{2} > \frac{1}{2}$ Now inductively we can prove $f$ is bijective for all $x$ just let $x,y$ be such that $f(x+y) - \frac{1}{2}$ is near $0$ so our bound is being divided by $2$ each time and goes from $\frac{1}{2}$ to $0$. Claim $:4 : f(kx) = f(k)f(x)$ Proof $:$ there exists $k'$ such that $f(k') = 2k-1$. $P(\frac{x}{2},\frac{x}{2},k') : \frac{k'+1}{2}f(x) = f(kx)$. $P(\frac{1}{2},\frac{1}{2},k') : k'+1 = 2f(\frac{f(k')+1}{2}) = 2f(k)$ so $\frac{k'+1}{2}f(x) = f(k)f(x)$ Claim $5 : f(\frac{1}{x}) = \frac{1}{f(x)}$ Proof $:$ Note that $f(1) = f(z)f(\frac{1}{z})$ so $f(\frac{1}{z}) = \frac{f(1)}{f(z)} = \frac{1}{f(z)}$ Claim $6 : f(f(z)) = z$. Proof $:$ from claim $3$ we have $f(f(z)+1) = \frac{z+1}{2}$ so $\frac{z+1}{2} = f(f(z)).f(1+\frac{1}{f(z)}) = f(f(z)).f(1+f(\frac{1}{z}))$ and Note that $f(1+f(\frac{1}{z})) = \frac{\frac{1}{z}+1}{2}$ so $\frac{z+1}{2} = f(f(z)) . \frac{\frac{1}{z}+1}{2} \implies f(f(z)) = z$. Claim $7 : f(2) = 2$. Proof $:$ Let $f(2) = c$. $f(f(2)) = f(c) = 2$. $P(1,1,z) : f(f(z)+1) = \frac{f(2)}{2}.(z+1)$ so $f(z) + 1 = f(\frac{c}{2}.(z+1)) = f(z+1).f(\frac{c}{2})$. Note that $f(c) = f(2).f(\frac{c}{2}) \implies f(\frac{c}{2}) = \frac{2}{c}$ so $f(z) + 1 = f(z+1).\frac{2}{c}$. put $z = 2$ and we have $c+1 = f(3).\frac{2}{c} \implies f(3) = (c+1)\frac{c}{2}$ so $f(4) = \frac{c}{2}((c+1)\frac{c}{2} + 1)$ and also $f(4) = f(2)^2 = c^2$ so $c^2 = \frac{c^3+c^2+2c}{4} \implies 4c = c^2 + c + 2 \implies c^2-3c+2 = 0 \implies c = 1$ or $c = 2$ and since $f(1) = 1$ then $f(2)$ can't be $1$ so is $2$. Also now we know that $f(z)+1 = f(z+1)$ Claim $8 : f(x+y) = f(x) + f(y)$. Proof $:$ $f(x+y) = f(x).f(1+\frac{y}{x}) = f(x).(1+f(\frac{y}{x})) = f(x) + f(y)$ Now $f$ is both additive and multiplicative so $f(x) = x$

$f$ is clearly injective $P(1,1,1) \Rightarrow f(2)=f(f(1)+1) \Rightarrow f(1)=1$ $P(x,x,z) \Rightarrow f(xf(z)+x)=\frac{1}{2}f(2x)(z+1)$ if we pin $x$ and slide $z$ along $\mathbb{R}^+$ we get that $(f(2x)/2, \infty) \subset Im(f)$ $\forall x \in \mathbb{R}^+$ $ (1)$ Let $f(\frac{1}{2})=c$ and $k\in \mathbb{R}^+$ $P(x,k-x,\frac{1}{2}) \Rightarrow \frac{3}{2}f(k)=f(xc+y)+f(yc+x) \Rightarrow min(f(xc+y),f(yc+x)) \leqslant \frac{3}{4} f(k)$ $\Rightarrow inf_{x\in \mathbb{R}^+}f(x) = 0$ $(2)$ $(1)+(2) \Rightarrow f $ is surjective $\Rightarrow f$ is bijective $P(\frac{1}{f(x)+1},\frac{1}{f(x)+1},x) * P(\frac{1}{2},\frac{1}{2},x) \Rightarrow f(\frac{2}{f(x)+1})f(\frac{f(x)+1}{2})=1$ and since $f$ is bijective $\Rightarrow f(x)f(\frac{1}{x})=1$ $\forall x \in \mathbb{R}^+$ $(*)$ Let $t\in \mathbb{R}^+$ such that $f(t)=3$ $P(\frac{1}{2},\frac{1}{2},t) \Rightarrow f(\frac{f(t)+1}{2})=\frac{t+1}{2}$ $(3)$ $P(\frac{x}{2},\frac{x}{2},t) \Rightarrow f(x)\frac{t+1}{2}=f(2x) \xRightarrow{(3)} f(2x)=f(x)f(2)$ $\forall x\in \mathbb{R}^+$ $(4)$ $f$ bijective $\Rightarrow \exists $ $a$ such that $f(a)=k$ $P(x,x,a) \xRightarrow{(3)} f(2x)f(\frac{k+1}{2})=f(x(k+1)) \xRightarrow{(4)} f(x)f(k+1)=f(x(k+1)) \Rightarrow f(x)f(y)=f(xy)$ $\forall x>0 $, $ y>1 \xRightarrow{(*)} f(x)f(y)=f(xy)$ $\forall x,y \in \mathbb{R}^+$ So $f$ is multiplicative $P(1,1,\frac{1}{x}) \Rightarrow f(f(\frac{1}{x})+1)=\frac{x+1}{2x}f(2)$ $P(1,1,x) \Rightarrow \frac{x+1}{2}f(2) = f(f(x)+1) = f(f(x))f(1+\frac{1}{f(x)}) = f(f(x))f(1+f(\frac{1}{x})) = f(f(x))\frac{x+1}{2x}f(2) \Rightarrow f(f(x))=f(x)$ So $f$ is an involution $(5)$ $f(f(x)+1)=\frac{(x+1)f(2)}{2} \xRightarrow{(5)} f(x)+1=f(\frac{(x+1)f(2)}{2})=f(x+1)f(f(2))\frac{1}{f(2)}=f(x+1)\frac{2}{f(2)}$ $\Rightarrow f(x+1)=(f(x)+1)\frac{f(2)}{2}$ $\frac{f(2)}{2}(f(3)+1)=f(4)=f(2)^2$ gives $f(2) = 2$ or $f(2) = 1$ but $f$ is injective $\Rightarrow f(2) = 2 \Rightarrow f(x+1)=f(x)+1$ $f(x+y)=f(y)(\frac{x}{y}+1)=f(y)(1+f(\frac{x}{y}))=f(y)+f(x)$ So $f$ is additive and multiplicative $\Rightarrow f = \text{Id}_{\mathbb{R}^+}$ (since $f(x)\equiv 0 $ is impossible)

shinichiman wrote: Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $$(z + 1)f(x + y) = f(xf(z) + y) + f(yf(z) + x),$$for all positive real numbers $x, y, z$. Fajar Yuliawan, Indonesia Let $P(x,y,z)$ be the assertion of the given functional equation. $P(x,x,z)\implies 2f(xf(z)+x)=(z+1)f(2x) \implies f\left(\frac{x}{2}f(z)+\frac{x}{2}\right)=\frac{(z+1)f(x)}{2},\quad\forall x,z>0.$ Hence $r\in \text{Im}(f)\implies \frac{3r}{4}\in \text{Im}(f)$, implying $f$ is surjective on $\mathbb R^+$ since $\lim \left(\frac{3}{4}\right)^n=0$. $P(1,1,z)\implies f(f(z)+1)=c(z+1)\implies f(cz+c+1)=c(f(z)+2),\quad\forall z>0,\text{ where }c=\frac{f(2)}{2},(*)$ $P(c+1,c+1,z)\implies f(cf(z)+f(z)+c+1)=b(z+1),\quad\forall z>0,\text{ where }b=\frac{f(2(c+1))}{2},$ so $$f(cf(z)+f(z)+c+1)=\frac{b}{c}f(f(z)+1)\implies f(cz+z+c+1)=\frac{b}{c}f(z+1),\quad\forall z>0$$by the surjectivity. Substitute $z$ by $z-1$ into the above FE, we get $$f((c+1)z)=\frac{b}{c}f(z),\quad\forall z>0.$$Substitute $z$ by $(c+1)z$ into $(*)$, we get $$\frac{b}{c}f(cz+1)=c\left(\frac{b}{c}f(z)+2\right)\implies f(cz+1)=cf(z)+\frac{2c^2}{b},\quad\forall z>0.$$Subtracting from the two sides of $(*)$ and the above FE, we get $$f(cz+c+1)-f(cz+1)=2c-\frac{2c^2}{b},\quad\forall z>0,$$so $$f(z+c)=f(z)+d,\quad\forall z>1,\text{ where } d=2c-\frac{2c^2}{b}.$$Comparing $P(x,y,z)$ and $P(x,y+c,z)$ with $y>1$, we get $$zd+f(yf(z)+x)=f(yf(z)+cf(z)+x),\quad\forall x,z>0,y>1.$$Let $x=1$ and $y\longrightarrow \frac{y-1}{f(z)}$, then the above FE becomes $$zd+f(y)=f(y+cf(z)),\quad\forall z>0,y>1,$$so $$f(y+cf(z+x))=zd+xd+f(y)=zd+f(y+cf(x))=f(y+cf(x)+cf(z)),\quad\forall x,z>0,y>1,$$but $f$ is easily injective, thus $f(x+z)=f(x)+f(z),\quad\forall x,z>0\implies f(x)=x,\quad\forall x>0.$

$\textit{Solution 1}$ Let $P(x, y, z)$ be the assertion above. $\textbf{Claim:}$ $f$ is bijective. $\textit{Proof}$ Notice $P(x, y, a)$ and $P(x, y, b)$ for $f(a) = f(b), a \neq b$, easily gives us a contradiction, giving $f$ is injective. For $f$ is surjective, notice $P(x, x, z)$ gives us \[(z+1)f(2x) = 2f(xf(z)+x)\]If $M = \inf\{f(x): x \in \mathbb{R^+}\}$, we get that $xf(z)+x$ has $2$ dimensions and can vary over all $\mathbb{R^+}$, giving a contradiction by setting $z$ arbitrarily large. Hence this gives us $f$ is surjective as desired. $\square$ If we consider $c$ such that $f(c) = 1$, taking $P(x, y, c)$ gives us that $c=1$ or $f(1) = 1$. $\textbf{Claim:}$ $f$ is multiplicative. $\textit{Proof}$ Now taking $P(\frac{1}{2}, \frac{1}{2}, z)$, we get \[f(\frac{f(z)+1}{2}) = \frac{z+1}{2}\]Now considering $P(\frac{x}{2}, \frac{x}{2}, z)$ \[f(\frac{xf(z)+x}{2}) = \frac{z+1}{2}f(x) = f(\frac{f(z)+1}{2})f(x)\]Pairing with the fact that $f$ is bijective, this gives us that $f$ is multiplicative where $f(ab) = f(a)f(b)$ and $b > \frac{1}{2}$. But notice if we consider $a < \frac{1}{2}, b > \frac{1}{2}$ and $ab > \frac{1}{2}$, we can take a $c < \frac{1}{2}$ to have \[f(abc) = f(ab)f(c) = f(a)f(b)f(c) = f(ac)f(b)\]Hence we get $f$ is multiplicative over all pairs of numbers. $\square$ $\textbf{Claim:}$ $f(3) = 3, f(2) = 2$. $\textit{Proof}$ Notice taking $P(\frac{1}{2}, \frac{1}{2}, z)$ again and $z \mapsto \frac{f(z)+1}{2}$, we get that \[f(\frac{z+3}{4}) = \frac{f(z)+3}{4}\]with $z \mapsto 4z+1$ again, to have \[4f(z+1) = f(4z+1) + 3\]for all $z \in \mathbb{R^+}$. Now taking $z=2$, we get $4f(3) = f(9) + 3 = f(3)^2 + 3$. This implies $f(3)^2 - 4f(3) + 3 = 0$, which immediately gives us that $f(3) = 3$ as $f$ is bijective. Now taking $P(\frac{1}{2}, \frac{1}{2}, 3)$ gives $f(2) = 2$. $\square$ To finish the problem, note $f(2^i) = 2^i$ for all $i \ge 1$ as $f$ is multiplicative. Also since $f\left(\frac{f(z)+1}{2}\right) = \frac{z+1}{2} \iff f(f(z)+1) = z+1$, we also have $f(2^{i}+1) = 2^i+1$ for all $i \ge 1$. Taking $P(x, y, 2^i)$ we get \[(2^i+1)f(x+y) = f(2^ix+y) + f(2^iy+x)\]where $(2^i+1)f(x+y) = f(2^i+1)f(x+y) = f((2^i+1)(x+y))$. If we choose the pair \[(x, y) = \left(\frac{2^ia-b}{2^{2i}-1}, \frac{2^ib-a}{2^{2i}-1}\right)\]where $a < b < 2^ia$, we have $f(a+b) = f(a)+f(b)$. But note since this applies for all $i \ge 1$, we have $f$ is additive over all pairs of values. This clearly gives $f(x) = x$. $\textit{Solution 2}$ We will assume the $f$ is bijective proof from Solution $1$. Now notice first taking $P(\frac{1}{2}, \frac{1}{2}, z)$, we have that Imf$(\mathbb{R^+}_{\geq \frac{1}{2}}) = \mathbb{R^+}_{\geq \frac{1}{2}}$. Since $f$ is bijective, we have that then $f$ must be bounded over the interval $(0, \frac{1}{2})$. Hence we know for any sequence $(x_n)$ such that $x_n \rightarrow 0$, we have that $\limsup_{n \rightarrow \infty} f(x_n)$ does indeed exist. Let this value be called $M$. Notice taking $x, y \rightarrow 0$ such that $x+y$ is equal to the limsup subsequence, we get that \[(z+1)M = f(xf(z)+y) + f(yf(z)+x)\]The RHS is bounded, but taking $z$ to be an arbitrarily large value gives us a contradiction. Hence $M = 0$. Thus we have $f \rightarrow 0$ as $x \rightarrow 0$. Now going back to $P(x, x, z)$, if we take $z \rightarrow 0$, we can see that $f$ actually must be continuous on all values $x$ as left-hand side changes value only by linear $\epsilon f(2x)$ and right-hand side has a change in input of $f(x+\delta) - f(x)$. Hence taking $z \rightarrow 0$ in original gives us Cauchy's Functional equation, concluding similarly to Solution $1$ that $f \equiv x$.

Very FUNctional equation, this is a different way from what i've done before (in the gsolve, which my friend fixed it later). Let $P(x,y,z)$ the assertion of the given functional equation. Claim 1: $f$ is bijective. Proof: Suppose $f(a)=f(b)$ then by $P(x,y,a)-P(x,y,b)$ we get that $a=b$ trivialy, now by $P(1,1,1)$ and injectivity we get that $f(1)=1$, so now by $P \left(\frac{1}{f(x)+1}, \frac{1}{f(x)+1}, x \right)$ $$f \left(\frac{2}{f(x)+1} \right)=\frac{2}{x+1} \implies f \; \text{surjective in} \; (0,2)$$And from $P(x,x,z)$ we get that $$f(x(f(z)+1))=\frac{z+1}{2} \cdot f(2x) \implies \; \text{if} \; m \in \; \text{Im}(f) \; \text{and} \; n>\frac{1}{2} \; \text{then} \; mn \in \; \text{Im}(f)$$And just to overkill this claim, by $P \left( \frac{1}{2}, \frac{1}{2}, x \right)$ we get that $$f \left(\frac{f(x)+1}{2} \right)=\frac{x+1}{2} \implies f \; \text{surjective in} \; \left(\frac{1}{2}, \infty \right)$$Hence $f$ is surjective and injective so its bijective as desired. Claim 2: $f$ is multiplicative. Proof: Let $t=\frac{f(z)+1}{2}$ then using Claim 1 we have $t$ takes all positive reals from $\frac{1}{2}$, so now by $P \left(\frac{x}{2}, \frac{x}{2}, z \right)$ we get that $$f(t)f(x)=f(xt) \; \forall x \in \mathbb R^+ \; \text{and} \; t>\frac{1}{2}$$Note it holds that $f \left( \frac{1}{t} \right)=\frac{1}{f(t)}$ due to how we defined $t$ and by $P \left(\frac{x}{f(z)+1}, \frac{x}{f(z)+1}, z \right)$ we get that $$f(x)f \left(\frac{1}{t} \right) =f \left(\frac{x}{t} \right) \; \forall t>\frac{1}{2} \; \implies f \; \text{multiplicative everywhere}$$Claim 3: $f(x)=x^n$ for some real $n$. Proof: By $P\left( f(x), f(x), \frac{1}{x} \right)$ and Claim 3 we get that $$f(f(x)+1)=\frac{x+1}{2x} \cdot f(2f(x)) \implies f(2f(x))=xf(2) \implies f(f(x))=x \implies f \; \text{involution}$$Now this gives us $f\left(\frac{x+1}{2} \right)=\frac{f(x)+1}{2}$ so if $a>\frac{1}{2}$ then $f(a)>\frac{1}{2}$, so if u let $g(x)=\ln(f(e^x))$ then $g$ is additive and also if $a>\frac{1}{2}$ then $g(a)=\ln(f(e^a))>-\ln(2)$ so $g$ is bounded hence its linear so $f(x)=x^n$ Finale: By $f(f(x))=x$ we get that $n=1$ or $n=-1$ for each $x$ but if for some $x$ it held that $f(x)=\frac{1}{x}$ then it holds that $f \left(\frac{x+1}{2x} \right)=\frac{x+1}{2}$ so either $x=1$ or $(x+1)^2=4x$, the last one also implying $x=1$ so $f(x)=x$ holds for every positive real $x$, thus we are done .

The answer is $f(x)=x$ only, which clearly works. Let $P(x,y,z)$ denote the assertion. Claim: $\inf f=0$. Proof: Let $M=\inf f$ and suppose $M>0$. Pick some $x+y$ with $f(x+y) \leq 1.1M$ and let $z=0.1$. Then $f(xf(z)+y)+f(yf(z)+x) \leq 1.21M$, which means we can't have both $f(xf(z)+y)$ and $f(yf(z)+x)$ be at least $M$: contradiction. $\blacksquare$ Claim: $f$ satisfies Jensen's. Proof: By comparing $P(x,y,z)$ and $P(\tfrac{x+y}{2},\tfrac{x+y}{2},z)$ we obtain $$f(xf(z)+y)+f(yf(z)+x)=2f\left(\frac{x+y}{2}(f(z)+1)\right).$$Now for any positive reals $a,b$, pick $f(z)<1$ to be sufficiently small such that $\tfrac{a+b}{f(z)+1}>\max(a,b)$. Then by putting $x=\tfrac{\tfrac{a+b}{f(z)+1}-a}{1-f(z)},y=\tfrac{\tfrac{a+b}{f(z)+1}-b}{1-f(z)}$, we find that $f(a)+f(b)=2f(\tfrac{a+b}{2})$. $\blacksquare$ Hence $f$ is linear, and it is straightforward to check that only $f(x)=x$ works. $\blacksquare$

IAmTheHazard wrote: By fixing $x=y$ and varying $z$ we find that $f$ is surjective. This is not enough to prove surjectivity. The domain is positive reals.

let P(x,y,z) be the statement above; computing p(0,0,z) we get that f(0)=0, then by computing p(x,y,0) we get that f(x+y)=f(x)+f(y) which is subdisfacted by all the functions in the form f(x)=cx, then assuming f(x)=k we get, (z+1)k=2k and so k=0 so either f(x)=cx or f(x)=0. Is it correct? Or did i gave something for granted and skipped a part of the proof? EDIT: by plugging in the value f(x)=cx i got that c must be equal to 1 so f(x) = x or f(x) = 0.

Marco22 wrote: let P(x,y,z) be the statement above; computing p(0,0,z) we get that f(0)=0, then by computing p(x,y,0) we get that f(x+y)=f(x)+f(y) which is subdisfacted by all the functions in the form f(x)=cx, then assuming f(x)=k we get, (z+1)k=2k and so k=0 so either f(x)=cx or f(x)=0. Is it correct? Or did i gave something for granted and skipped a part of the proof? EDIT: by plugging in the value f(x)=cx i got that c must be equal to 1 so f(x) = x or f(x) = 0. 0 is not in the domain

tadpoleloop wrote: IAmTheHazard wrote: By fixing $x=y$ and varying $z$ we find that $f$ is surjective. This is not enough to prove surjectivity. The domain is positive reals. Oops Edit: should be fixed?

Let $x = y$, then $\frac{z+1}{2} f(2x) = f(x(f(z)+1))$. Thus the range of $f$ includes all intervals $\bigcup \left(\frac{f(2x)}{2^k}, \infty \right)$, so $f$ is surjective. Let $c$ be any value such that $f(c) = 1$. Then $(c+1)f(x+y) = 2f(x+y)$ using $z = c$, so $c = 1$. Now choose any $a, b > 2$. Then it follows, choose $z$ so that $f(z) = a + b - 1 \geq 3$. Then $1 + f(z) = a + b = 2 + (f(z) - 1)$. Now since $2, f(z) - 1, a, b \in (1, f(z))$ it follows that there exist $x_1, y_1, x_2, y_2$ so that $x_1 + y_1 = x_2 + y_2 = 1$ and plugging $x_1, y_1$ or $x_2, y_2$: $z+1 = f(2) + f(a+b-2) = f(a) + f(b)$. Thus since $a + b - 2 = 2 + (a-2) + (b-2) > 2$, it follows that a shifted version of $f$ satisfies Cauchy's functional equation on $(0, \infty)$ and maps positive reals to reals bounded below by $-f(2)$. In particular, it follows that $f$ is never negative, so $f$ is monotonic increasing, so $f$ is linear (the proof proceeds identically to the proof that Cauchy + bounded implies linear). Now it follows that $f = ax + b$ for $x > 2$. It follows from plugging large $z, x, y$ that $f = x$ for $x > 2$. Now reconsider $(z+1)f(2x) = 2f(x(f(z)+1))$ for any $x$. Sending $z$ to infinity, it follows that $f(2x) = 2x$, so we have $f(x) = x$ for all $x$.

Denote $P(x,y,z)$ as the assertion of the equation in the problem.

Since $f$ is additive over real positive, then $f(x)=ax$ for some constant $a\in \mathbb{R^+}$. From claim 4, we get $a=1$. Hence, $f(x)=x \ \forall \ x \in \mathbb{R^+}.$

\[(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x)\]Only function holds the conditions is $f(x)=x$. Let $P(x,y,z)$ be the assertion. Claim: $f$ is injective. Proof: Comparing $P(f(x),1,z)$ with $P(f(z),1,x)$ gives \[ (z+1)f(f(x)+1)=f(f(x)f(z)+1)+f(f(x)+f(z))=(x+1)f(f(z)+1)\]Choose $z=1$ to verify $2f(f(x)+1)=(x+1)f(f(1)+1)$. Let $f(f(1)+1)=2c$. We observe that $f(f(x)+1)=(x+1)c$ which proves the injectivity.$\square$ Claim: $f(1)=1$. Proof: $P(1,1,z)$ gives $f(2)(z+1)=2f(f(z)+1)=2c(z+1)$. Thus, $f(2)=2c=f(f(1)+1)$. Since $f$ is injective, we conclude that $f(1)=1$.$\square$ Claim: $f(\frac{f(x)+1}{2})=\frac{x+1}{2}$. Proof: $P(\frac{1}{2},\frac{1}{2},z)$ yields $z+1=2f(\frac{f(z)+1}{2})$.$\square$ Claim: $c=1$. Proof: \[P(x,x,f(2z+1)+2): \ \ (f(2z+1)+3)f(2x)=4f(x(z+2))\]Take $x=z=1$ to get $(f(3)+3)c=2f(3)\iff f(3)=\frac{3c}{2-c}$. Note that this means $c<2$. Now by using $f(f(x)+1)=(x+1)c$, \[f(f(\frac{c+1}{2-c})+1)=c(\frac{c+1}{2-c}+1)=\frac{3c}{2-c}=f(3)\implies f(\frac{c+1}{2-c})=2\]\[P(\frac{c+1}{2(2-c)},\frac{c+1}{2(2-c)},z): \ \ f(\frac{f(2z+1)+1}{2})=z+1=f((f(z)+1)(\frac{c+1}{2(2-c)}))\]Hence $f(2z+1)+1=(f(z)+1)(\frac{c+1}{2-c})$. Plugging $z=2$ gives $f(5)=\frac{(2c+1)(c+1)}{2-c}-1$. \[P(\frac{1}{2},\frac{1}{2},\frac{f(5)+1}{2}): \ \ 2c=f(2)=\frac{f(5)+3}{4}\]\[\frac{(2c+1)(c+1)}{2-c}-1=f(5)=8c-3\implies 2c^2+3c+1-2+c=16c-6-8c^2+3c\iff (c-1)^2=0\iff c=1\]Which completes the proof.$\square$ Claim: $f$ is additive. Proof: We have $f(f(x)+1)=x+1$. \[P(x,y,f(z)+1): \ \ (f(z)+2)f(x+y)=f(xz+x+y)+f(yz+x+y)\]Plugging $x=y=1$ yields $f(z)+2=f(z+2)$. Note that by induction, $f(k)=k$ for all positive integers. $P(3,1,f(z)+1)$ implies $4(f(z)+2)=f(3z+4)+f(z+4)=f(3z)+f(z)+8$ hence $3f(z)=f(3z)$. \[P(x,x,2): \ \ 3f(2x)=2f(3x)=6f(x)\implies f(2x)=2f(x)\]$P(x,x,f(z)+1)$ gives $2f(z+2)f(x)=(f(z)+2)f(2x)=2f(xz+2x)$ thus, $f(x)f(z+2)=f(x(z+2))$. So \[f(xz+yz+2x+2y)=f(z+2)f(x+y)=f(xz+x+y)+f(yz+x+y)\]We claim that for all $a,b\in R^+$, we can find $x,y,z$ such that $xz+x+y=a$ and $yz+x+y=b$. For $a=b,$ we proved that $f(2a)=2f(a)$. We can assume that $a>b$. \[b-x-yz=y=a-x-xz\implies z=\frac{a-b}{x-y}\]$\frac{y(a-b)}{x-y}+x+y=b\iff ya-by+x^2-y^2=bx-by\iff x^2-bx+(ya-y^2)=0$. We choose $y<\min\{a,b/2\}$. \[x=\frac{b-\sqrt{4y^2-4ya+b^2}}{2}>\frac{b-\sqrt{(b-2y)^2}}{2}=y\]So $z>0$. Since there exists $x,y,z$ for all $a,b$ we observe that $f(a)+f(b)=f(a+b)$ hence $f$ is additive.$\square$ \[(z+1)f(x)+(z+1)f(y)=f(xf(z))+f(yf(z))+f(x)+f(y)\iff zf(x)+zf(y)=f(xf(z))+f(yf(z))\]Plug $P(x+2,y,f(z)+1)$ to get (by using $f(x)f(y+2)=f(xy+2x))$ \[(f(z)+1)f(x+2)+(f(z)+1)f(y)=f(xz+x+2z+2)+f(yz+y)\]\[2f(z)+f(x)+f(y)f(z)+f(y)=f(x)+2f(z)+f(yz)+f(y)\iff f(y)f(z)=f(yz)\]Thus, $f$ is multiplicative. Since $f$ is both additive and multiplicative, we get that $f(x)=x$ as desired.$\blacksquare$

Why the minnn