Allnames wrote: -Find all the functions from $ \mathbb R^ + \to \mathbb R^ +$ that satisfy: $ x^2(f(x) + f(y)) = (x + y)(f(yf(x)))$ From condition we have : $ \frac {f(yf(x))}{x^2} = \frac {f(xf(y))}{y^2}$(1) Now take $ x\to y$ then we have : \[ f(xf(x)) = xf(x) \] Therefore f has a fixed point $ a$ (i.e $ f(a) = a$) From $ f(xf(x)) = xf(x)$ we have $ f(a^n) = a^n,\forall n\in N$ Take $ x = a^m,y = a^n$ into (1) we have $ a^{2m} = a^{2n}\Rightarrow a = 1$ It means that f has only a fixed point is 1 . So $ xf(x) = 1$ ,it gives $ f(x) = \frac {1}{x}$ Easy to check that this function satisfy condition .

Nice,TT sphn,and here is my proof If $ f$ is const we have $ f(x)=0$,wrong If not,Fix $ y$ we can easy find $ f(x_1)=f(x_2)$<=>$ x_1=x_2$ Let $ x=y=1$ we have $ f(f(1))=f(1)$ thus$ f(1)=1$ Let $ x=1$ we have $ yf(y)=1$ hence $ f(y)=\frac{1}{y}$done

Suppose $f(a)=f(b)$ for distinct $a,b$. we have $\frac{a^2}{a+y}=\frac{b^2}{b+y}$, we get $ab=-y(a+b)$ impossible, $a,b,y>0$, so $a=b$, so f is injective.Take $x=y=1$ and we have $f(1)=f(f(1)),$ so$ f(1)=1$ Taking $x=1$,we have $yf(y)=1$ so the conclusion follows.

let $f(1)=c$ let $x=1,y=x$,then$f(x)+c=(1+x)f(cx)$. let $y=1$,them $x^2(f(x)+c)=(x+1)f(f(x))$ hence $f(f(x))=x^2f(cx)$ let $x=c$,then$c^2f(c^2)=f(f(c))$ let $x=y$ we get $xf(x)=f(xf(x))$ let $x=1$,$f(c)=c$. let $x=c$,$f(cf(c))=cf(c)$,yielding $f(c^2)=c^2$ hence $c^2=\frac{f(f(c))}{c^2}=\frac{1}{c}$ ,$c=1$. hence $f(x)+1=(1+x)f(x)$,obtaining $f(x)=\frac{1}{x}$ QED

Ans: $f(x)=\frac{1}{x}$. Sol: Let $P(x,y)$ be the given assertion, \[P(x,x)\implies f(xf(x))=xf(x).\]Let $Q=xf(x)$, then \[P(Q,1)\implies Q^3+Q^2f(1)=(Q+1)Q \implies Q^2+(f(1)-1)Q-1=0\]this is a quadratic equation in $Q$ and since $Q\ne 0$, we also have the discriminant being positive, so $Q$ must be constant which means \[xf(x)=C \implies f(x)=\frac{C}{x}\]for some positive real $C$. Plugging into our original equation yields $f(x)=\frac{1}{x}$ and it is easy to see that this satisfies our equation.$\square$

fakesolve

bump, read @above

@above: just notice. $$f(xf(x))=xf(x)$$Now plug $x\to xf(x)$ and we have \begin{align*} f(xf(x)f(xf(x)))&=xf(x)f(xf(x))\\ f(xf(x)xf(x))&=xf(x)xf(x)\\ f((xf(x))^2)&=(xf(x))^2 \end{align*}So by induction letting $a=xf(x)$ we have $f(a^k)=a^k$. Now we take $(m,k)\in\mathbb R^2$ for $m\neq k$ s.t $x=a^m$ and $y=a^k$. Using the relation we get $a^{2m}=a^{2k}$. Now say $k$ is fixed value. Consider $g(x)=a^x$. So $g(2m)=g(2k)\forall m$ so $g$ is constant which can be only possible for $a=1$. Hence $a=1$ and $xf(x)=a=1$.

We plug in $x=y$ from here we get that $xf(x) = f(xf(x))$. We define set $S$ to be $S= \{ xf(x) \mid \forall x \in \mathbb{R}^+ \}$, from the above we have that if $x \in S$ then we have that $f(x)=x$. Now we plug in $x,y$ such that they are in $S$ this gives us $x^2=f(yf(x))=f(yx)=f(xf(y))=y^2$. This gives us that $x=y$, which means that $xf(x)$ is a constant value. By direct computation we get that the only solution is $f(x)=\frac{1}{x}$

Nice problem. Let $P(x,y)$ be the assertion of $x^2(f(x)+f(y))=(x+y)f(yf(x))$. $P(x,y)\implies x^2(f(x)+f(y))=(x+y)f(yf(x))$ $P(y,x)\implies y^2(f(x)+f(y))=(x+y)f(xf(y))$ Thus, $f(yf(x))\cdot y^2=f(xf(y))\cdot x^2$ Setting $y=1$, we have that $f(f(x))=f(xf(1))\cdot x^2$. $P(1,1)\implies f(1)=f(f(1))$ $P(1,x)\implies f(x)+f(1)=(x+1)f(xf(1))\quad \heartsuit$ $P(x,1)\implies x^2(f(x)+f(1))=(x+1)f(f(x))=(x+1)f(xf(1))\cdot x^2$ Thus, $f(x)+f(1)=(x+1)f(xf(1))\quad \diamondsuit$ $f(1)=f(f(f(1)))=f(f(1)^2)\cdot f(1)^2\implies f(f(1)^2)\cdot f(1)=1$. Substituting $x=f(1)$ to $\diamondsuit$, we get that $f(f(1))+f(1)=(f(1)+1)f(f(1)^2)\implies 2f(1)^2=f(1)+1\implies 2f(1)^2-f(1)-1=0$ Factorizing, we get that $(f(1)-1)(2f(1)+1)=0$, thus $f(1)=1$. Thus by $\heartsuit$, $f(x)+1=(x+1)f(x)\implies f(x)=\frac{1}{x}$. This does work. Answer. $f(x)=\frac{1}{x}\,\forall x\in \mathbb R^+$.

Allnames wrote: -Find all the functions from $ \mathbb R^+ \to \mathbb R^+$ that satisfy: $ x^2(f(x)+f(y))=(x+y)(f(yf(x)))$ Nice problem. $P(x,x)\implies f(xf(x))=xf(x)$ $P(xf(x),yf(y))\implies f(xyf(x)f(y))=(xf(x))^2$ $P(yf(y),xf(x))\implies f(xyf(x)f(y))=(yf(y))^2$ Then $(xf(x))^2=(yf(y))^2\implies f(x)=\frac{c}{x},c>0$ $f(xf(x))=xf(x)\implies \frac{c}{x\cdot \frac{c}{x}}=x\cdot \frac{c}{x}\implies c=1$ Then $f(x)=\frac{1}{x}$

Let $P(x,y)$ be the assertion $x^2f(x)+x^2f(y)=xf(yf(x))+yf(yf(x))$. $P(1,1)\Rightarrow f(f(1))=f(1)$ $P(f(1),1)\Rightarrow f(1)=1$ Switching $x,y$ we have: $\frac{P(y,x)}{P(x,y)}\Rightarrow x^2f(xf(y))=y^2f(yf(x))$ Taking $y=1$, it's $f(f(x))=x^2f(x)$. $P(x,1)\Rightarrow\boxed{f(x)=\frac1x}$ which works.

Let $P(x,y)$ be the assertion of the problem *Claim:* there are two positive real numbers $a,b$ such that $f(a)=f(b)=c$ $P(a,y): a^2(c+f(y))=(a+y)(f(yc))\implies \dfrac {c+f(y)}{f(yc)}=\dfrac {a+y}{a^2}$ $P(b,y): \dfrac {c+f(y)}{f(yc)}=\dfrac {b+y}{b^2}$ So $\dfrac {b+y}{b^2}=\dfrac {a+y}{a^2}$ hence $a=b$ so $f$ is injective $P(1,1): f(f(1))=f(1)\implies f(1)=1$ $P(1,x): 1+f(x)=(1+x)f(x)$ Finally $\boxed {f(x)=\dfrac {1}{x} \forall x\in \mathbb R^+}$

Denote the assertion by $P(x,y).$ If $f(u)=f(v)$ then $P(u,z)$ and $P(v,z)$ gives $u^2(v+z)=v^2(u+z)$ so $u=v.$ $P(1,1)\implies f(1)=1.$ $P(1,x)\implies f(x)=x^{-1}$ this fits.