sororak wrote: Let $\mathbb{R}^{+}$ be the set of positive real numbers. Find all functions $f:\mathbb{R}^{+}\to \mathbb{R}^{+}$ such that for all positive real numbers $x,y$ the equation holds: $(x+y)f(f(x)y)=x^2f(f(x)+f(y))$ Let $P(x,y)$ be the assertion $(x+y)f(f(x)y)=x^2f(f(x)+f(y))$ If $f(a)=f(b)$, then comparing $P(a,x)$ and $P(b,x)$ implies $\frac{x+a}{a^2}=\frac{x+b}{b^2}$ $\forall x>0$ and so $a=b$ and $f(x)$ is injective. $P(\frac 32,\frac 34)$ $\implies$ $f(\frac 34f(\frac 32))=f(f(\frac 32)+f(\frac 34))$ and, since injective : $\frac 34f(\frac 32)=f(\frac 32)+f(\frac 34)$ which implies $f(\frac 34)+\frac 14f(\frac 32)=0$, which is impossible since $f(x)$ is from $\mathbb R^+\to\mathbb R^+$ So no solution to this equation.

my solution obviously,if $f(a)=f(b)$ then we have $a=b$,let $y=x^{2}-x$,here $x>1$,then we have $f((x^{2}-x)f(x))=f(f(x)+f(x^{2}-x))$ hences to $(x^{2}-x-1)f(x)=f(x^{2}-x)$,let x be the larger root of $x^{2}-x-1=0$ a contradiction!

Does this work? Let $P(x,y)$ be the assertion $(x+y)f(f(x)y)=x^2 f(f(x)+f(y)).$ $\exists a,b\in\mathbb{R}^{+},$ such that $f(a)=f(b),$ comparing $P(a,x)$ and $P(b,x),$ we find $\frac{a^2}{a+x}=\frac{b^2}{b+x}\to (a-b)(ab+ax+bx)=0\to a=b,$ since $ab+ax+bx>0.$ Then $f$ is injective. From $P(x,y)$ and $P(y,x)$ we find $$f(f(y)x)=f(f(x)y)\to ^{injectivity} \frac{f(y)}{y}=\frac{f(x)}{x}=a,$$where $a\in\mathbb{R}^{+}.$ But in our functional equation we replace $f(x)=ax$ we find $x^2=xy$ but this is not true for infinitely $x,y.$ Them no solution to this fe.

Ferid.---. wrote: . From $P(x,y)$ and $P(y,x)$ we find $$f(f(y)x)=f(f(x)y)\to ^{injectivity} \frac{f(y)}{y}=\frac{f(x)}{x}=a,$$where $a\in\mathbb{R}^{+}.$ I guess you've made a mistake here. From $P(x,y)$ and $P(y,x)$ we get:$$f(yf(x))x^2=f(xf(y))y^2$$

m.yekta wrote: Ferid.---. wrote: . From $P(x,y)$ and $P(y,x)$ we find $$f(f(y)x)=f(f(x)y)\to ^{injectivity} \frac{f(y)}{y}=\frac{f(x)}{x}=a,$$where $a\in\mathbb{R}^{+}.$ I guess you've made a mistake here. From $P(x,y)$ and $P(y,x)$ we get:$$f(yf(x))x^2=f(xf(y))y^2$$ Thank you.

dame dame

Let for some $a\neq b$, we have $f(a) = f(b)$. Then comparing $P(a,y)$ and $P(b,y)$ gives $a=b$, so $f$ is injective. $P(x,x^2 - x)$ for $x > 1$ gives us $f(f(x)(x^2 - x)) = f(f(x) + f(x^2 - x)) \implies f(x)(x^2 - x - 1) = f(x^2 - x)$ for all $x > 1$. So, plugging $x = \sqrt{2}$ we have $f(\sqrt{2})(1 - \sqrt{2}) = f(2 - \sqrt{2}) < 0$, contradiction. So there is no function satisfying the condition.

f is injective(easy) Exist $a$ positive real, such that $$a^2=a+1$$$P(a,1) \implies \boxed{f(1)=0}$ Contradiction.

$P(x,f(y)) : \frac{x+f(y)}{x^2} = \frac{f(f(x) + f(f(y)))}{f(f(x)f(y))}$. Let Assume $f(a_1) = f(a_2) = c$. $P(a_1,f(a_2)) , P(a_2,f(a_1)) : \frac{a_1+c}{{a_1}^2} = \frac{a_2+c}{{a_2}^2} \implies a_1 = a_2$ so $f$ is injective. $P(x,-x) , P(0,y) : f(f(x) + f(-x)) = f(f(0)y) = 0 \implies f(0)y = f(x) + f(-x)$. But Now for $x = 0$ we have $y = 2$ which is not true because $y$ is not constant so no such $f$ exists.

Lets speedrun. Let $P(x,y)$ the assertion of the F.E. then if $f(a)=f(b)$ then by comparing $P(a,x)$ and $P(b,x)$ u get injectivity. Now by $P(\phi,1)$ where $\phi$ is the golden ratio, we have that $f(1)=0$ which is impossible so no much function exists.