KPBY0507 wrote: Find all functions $f:R^+\rightarrow R^+$ such that for all positive reals $x$ and $y$ $$4f(x+yf(x))=f(x)f(2y)$$ Let $P(x,y)$ be the assertion $4f(x+yf(x))=f(x)f(2y)$ 1) there are no injective solution Comparing $P(2x,y)$ with $P(2y,x)$, we get $f(2x+yf(2x))=f(2y+xf(2y))$ and, if $f(x)$ is injective : $2x+yf(2x)=2y+xf(2y)$, which is $\frac{f(2y)-2}{y}=\frac{f(2x)-2}x$ And so $f(x)=ax+2$ for some constant $a\ne 0$ Which is never a solution. Q.E.D. 2) The only non injective solution is $f\equiv 4$ 2.1) $\exists u$ such that $f(u)=4$ Since non injective, let $f(a)=f(a+\Delta)$. $P(a,\frac {\Delta}{f(a)})$ $\implies$ $f(\frac{2\Delta}{f(a)})=4$ Q.E.D. 2.2) $f(2x+u)=f(x)\quad\forall x$ This is just $P(u,\frac x2)$ Q.E.D. 2.3) $\boxed{f(x)=4\quad\forall x>0}$ Comparing $P(x,y)$ with $P(2x+u,y)$, we get $f(x+yf(x))=f(2x+u+yf(2x+u))$ Moving there $y\to \frac y{f(x)}$, we get $f(x+y)=f(2x+u+y)$ $\forall x,y$ Let $t>0$ and $x\in(t+u,2t+u)$ Plugging $(x,y)=(x-t-u,2t+u-x)$ in $f(x+y)=f(2x+u+y)$, we get $f(t)=f(x)$ And so $f(x)$ is constant over $(t+u,2t+u)$ $\forall t>0$ And so $f(x)$ is constant over $(u,+\infty)$ (just use as many overlapping intervals as needed). Plugging this in $P(x,y)$, this is $f(x)=4$ $\forall x>u$ Then, choosing $x>u$, $P(x,y)$ becomes $f(2y)=4$ $\forall y$, which indeed fits. Q.E.D.

case(1):$f$ is non-injective let $f(a)=f(b)$ $P(a,\frac{y}{f(a)})$ with $P(b,\frac{y}{f(a)})$ we'll get $$f(y+c)=f(y):c=a-b>0$$$P(x,y+c) \implies f(z+cf(x))=f(z)$ for any large $z$ $P(z,c) \implies f(2c)=4=f(c)$ $P(c,y) \implies f(2y)=f(y)$ $$4f(x+yf(x))=f(x)f(2y)=f(x)f(2^ny) : \forall n \in \mathbb{N}$$choose $n$ such that $2^n>f(x)$ $P(x,\frac{x}{2^n-f(x)}) \implies f(x)=4 \forall x \in \mathbb{R}^+$ case(2): $f$ is injective switching $x,y$ in $P(2x,y) $ gives that there's no such $f$

Ali3085 wrote: case(1):$f$ is non-injective let $f(a)=f(b)$ $P(a,\frac{y}{f(a)})$ with $P(b,\frac{y}{f(a)})$ we'll get $$f(y+c)=f(y):c=a-b>0$$$P(x,y+c) \implies f(z+cf(x))=f(z)$ for any large $z$ $P(z,c) \implies f(2c)=4=f(c)$ $P(c,y) \implies f(2y)=f(y)$ $$4f(x+yf(x))=f(x)f(2y)=f(x)f(2^ny) : \forall n \in \mathbb{N}$$choose $n$ such that $2^n>f(x)$ $P(x,\frac{x}{2^n-f(x)}) \implies f(x)=4 \forall x \in \mathbb{R}^+$ case(2): $f$ is injective switching $x,y$ in $P(2x,y) $ gives that there's no such $f$ and we win

This problem is actually the same with 2005 ISL A2... https://artofproblemsolving.com/community/c6h85071p494821

My solution is a little bit different. Claim 1: For any $x{}$, $f(x)\geq 4$. Proof: First of all, assume that for some $x$, $f(x)<2$. Then, take $y=\frac{x}{2-f(x)}$. This gives us \[4f\bigg(\frac{2x}{2-f(x)}\bigg)=4f\bigg(x+\frac{x}{2-f(x)}\cdot f(x)\bigg)=4f(x)f\bigg(\frac{2x}{2-f(x)}\bigg)\]so $f(x)=4$, which is a contradiction, since we assumed $f(x)<2$. Therefore, for any $x$, $f(x)\geq 2$. Therefore, we have \[f(x)^2=4f\big(x+\frac{x}{2}f(x)\big)\geq 8\]which gives us $f(x)\geq 2^{\frac{3}{2}}$ for any $x$. Using this again, we have \[f(x)^2=4f\big(x+\frac{x}{2}f(x)\big)\geq 4\cdot 2^{\frac{3}{2}}=2^{\frac{7}{2}}\]so for any $x, \ f(x)\geq 2^{\frac{7}{4}}$. Inductively, one can prove that for any $n, \ f(x)\geq 2^{\frac{2^{n+1}-1}{2^n}}$ so for any $x{}$ \[f(x)\geq \lim_{n\to\infty} 2^{\frac{2^{n+1}-1}{2^n}}=4.\]Claim 2: The function $f$ is non-decreasing. Proof: Using claim $1$ and the condition in the statement, we know that \[4f(x+yf(x))=f(x)f(2y)\geq 4f(x)\Longrightarrow f(x+yf(x))\geq f(x).\]Take $y\to 0$ and we get that for any $\varepsilon>0,$ $f(x+\varepsilon)\geq f(x)$, so we are done. Claim 3: There exist $a$ and $b$ such that $a\neq b$ and $f(a)=f(b)$ (i.e. $f$ is not injective). We will also find explicit values for $a,b$ Proof: Observe that \[4f(2x+yf(2x))=f(2x)f(2y)=4f(2y+xf(2y))).\]if $f$ were injective, then we must have \[2x+yf(2x)=2y+xf(2y)\geq 2y+4x\text{ (using claim }1\text{)}\]so we have \[y\geq\frac{2x}{f(2x)-2}\]which is an obvious contradiction as we can take $y\to 0$. So $f$ cannot be injective and we found our $a{}$ and $b{}$. We are now ready to prove that the only function which satisfies the condition in the statement is $f\equiv 4$. Using claim 3, consider $a$ and $b$ such that $f(a)=f(b)$. The function is increasing, so $f$ is constant on inbetween $a{}$ and $b{}$. We know that $a=2y+xf(2y)$ and $b=2x+yf(2x)$ for some arbitrary $x{}$ and $y\to 0$. Note that because $y\to 0$ then $a\to 4x$ (the limit in 0 exists and it is equal to 4) and $b\to 2x$. Therefore, $f$ is constant on $[2x,4x]$ for any $x{}$ so $f{}$ is constant on $[2,\infty)$. It then follows easily that $f\equiv 4$.

KPBY0507 wrote: Find all functions $f:R^+\rightarrow R^+$ such that for all positive reals $x$ and $y$ $$4f(x+yf(x))=f(x)f(2y)$$ Solved with Rama1728 Let $P(x,y)$ the assertion of the F.E. as always lmao Claim 1: $f(x) \ge 4$ for every $x$. Proof: First if there exists $x$ with $f(x)<2$ then by $P \left(x, \frac{x}{2-f(x)} \right)$ $$f(x)=2 \; \text{contradiction!!} \; \implies f(x) \ge 2$$Now with $P \left(x, \frac{x}{2} \right)$ $$f(x)^2 \ge 8 \implies f(x) \ge 2^{\frac{3}{2}}$$And using that again we get $f(x) \ge 2^{\frac{7}{4}}$, in general if we do it infinitly times we will end up with $f(x) \ge 4$ which is easy to verify, thus we are done Claim 2: $f$ is linear. Let $y$ to be smaller enough on the F.E. and using Claim 1 we get $$f(x+yf(x)) \ge f(x) \implies f \; \text{increasing}$$Since $f$ is increasing now using Lebesgue monotone diferentiable theorem we have that there exists $x_0$ such that $f'(x_0)$ exists, hence by $P(x_0,y)$ and limit of $y$ going to $0$ $$\lim_{y \to 0^+} \frac{f(2y)-4}{4y}=c \; \text{exists}$$Now since that limit doesnt depend of $x$ we get by $P(x,y)$ and limit of $y$ going to $0$ $$f'(x)=c \implies f(x)=ax+b$$Thus we are done Final proof: Using the existence of $\lim_{y \to 0^+} \frac{f(2y)-4}{4y}$ we can easly get $b=4$ and now we bash for getting $a$. $$4ax+4a^2xy+16a+16=2a^2xy+4ax+8ay+16 \implies 2a^2xy=8ay-16a \implies a=0$$Hence $f(x)=4$ works and we are done

Feels like 2005 A2 KPBY0507 wrote: Find all functions $f:R^+\rightarrow R^+$ such that for all positive reals $x$ and $y$ $$4f(x+yf(x))=f(x)f(2y)$$ Note that if $f(x)<2$ for any $x$; plugging $y=\frac{x}{2-f(x)}$ implies $f(x)=4$, a contradiction! Thus $f(x) \geqslant a_0=2$ for all $x$. Now plugging $y=x/2$ and noting $f(t) \geqslant a_n$ for all $t$ implies $f(x)\geqslant \sqrt{4a_n}=a_{n+1}$ for all $x$; where $a_{n+1}=2\sqrt{a_n}$ for all $n \geqslant 0$. Thus, $f(x) \geqslant \text{lim}_{n \rightarrow \infty} a_n=4$ for all $x$. Further, assuming $f$ is not injective, say $f(u)=f(v)$ with $u<v$, plugging $x=u$ and $x=v$ gives us that $f(z)=f(z+(v-u))$ for all $z \geqslant u$. However, $f(2y) \geqslant 4$ in the equation implies $f$ is increasing, so together with the above, we get that for all $w \geqslant u$, $f(u) \leqslant f(w) \leqslant f(u+k(v-u))=f(u)$ for some sufficiently large $k \in \mathbb{N}$, hence $f$ is constant on $[u, \infty)$, and hence $f \equiv 4$ on this interval by plugging any large $x, y$. Pick $x<u$ and make $y$ large, to get $f(x)=4$ as well, hence $f \equiv 4$. Now assuming $f$ is injective, we have $f(2x+yf(2x))=f(2y+xf(2y))=\tfrac{1}{4}f(2x)f(2y)$ hence $x(2-f(2y))=y(2-f(2x))$, so $\tfrac{x}{2-f(2x)}$ is a constant for all $x$; solving which gives $f(t)=2+2tc$ for some constant $c$; which does not yield any solutions to the given equation. Thus $f \equiv 4$ is the only solution, and clearly it works.

KPBY0507 wrote: Find all functions $f:R^+\rightarrow R^+$ such that for all positive reals $x$ and $y$ $$4f(x+yf(x))=f(x)f(2y)$$ Let $P(x,y)$ be the assertion $4f(x+yf(x))=f(x)f(2y)$. Assuming $f$ is injective, by $P(2x,y)$ and symmetry we have $$f(2x)f(2y) = 4f(2x+yf(2x))= 4f(2y+xf(2y)) \rightarrow 2x+yf(2x) = 2y+xf(2y) \rightarrow \dfrac{f(2x)-2}{x} = \dfrac{f(2y)-2}{y}.$$So in this case $f(x) = cx+2$ for all $x>0$ which doesn't work. If there exists $a,b>0$ such that $f(a)=f(b)$ then by $P(a,y/f(a)), P(b,y/f(b))$ we get that $f(y+a) = f(y+b)$. So $f(y)=f(y+t)$ for all $y>b$ where $a>b$ are such that $f(a)=f(b)$. Let this phrase be the assertion $Q(y,a,b)$. If $f(a)=f(b)$ for $a>b$ then $P(b, \dfrac{a-b}{f(a)})$ makes $x = b, a= x+yf(x)$ and give us that there is $t>0$ such that $f(t)=4$. By $P(t,x/2)$ we get $$f(x) = f(2x+t) \; \forall x>0.$$We can now use $Q(y,2x+t,x)$ which gives us $$\forall x>0 , y>x :\;\; f(y)=f(y+x+t)$$and this means that $f$ is eventually constant. i.e there is $N,c>0$ such that $f(x) = c$ for all $x>N$. By choosing $y$ large enough in $P(x,y)$ we have $cf(x)= 4c$ which means $f(x)=4$ for all $x>0$. This is the only solution.