cpn wrote: Find all function $ f:\mathbb R \to \mathbb R$ such that $xf(y)+yf(x)=(x+y)f(x)f(y)$ Put $y:=1$ we have $f(x)[xf(1)+f(1)-1]=xf(1)$ $x=y=1$ we have $f(1)=(f(1))^2 $ If $f(1)=0 \Rightarrow f(x)=0$ If $f(1)=1 \Rightarrow f(x)=1$ So $f(x)=0$ and $f(x)=1$ $\forall x \in \mathbb R$

cpn wrote: Find all function $ f:\mathbb R \to \mathbb R$ such that $xf(y)+yf(x)=(x+y)f(x)f(y)$ Let $P(x,y)$ be the assertion $xf(y)+yf(x)=(x+y)f(x)f(y)$ $f(x)=0$ $\forall x$ is a solution and let us from now look onlu for non all-zero solutions. 1) if $f(0)\ne 0$ : $P(x,0)$ $\implies$ $xf(0)=xf(x)f(0)$ and so $f(x)=1$ $\forall x\ne 0$ And the function $f(0)=a$ and $f(x)=1$ $\forall x\ne 0$ is indeed a solution. 2) $f(0)=0$ and $f(u)\ne 0$ for some $u\ne 0$ If $f(y)=0$ for some $y\ne 0$, $P(u,y)$ becomes $yf(u)=0$ and so contradiction. So $f(x)=0$ $\iff$ $x=0$ Let then $x,y\ne 0$ and $g(x)=\frac x{f(x)}$. $P(x,1)$ becomes : $g(x)=x+1-g(1)$ Setting $x=1$ in this equation, we get $g(1)=1$ and $g(x)=x$ and $f(x)=1$ which indeed is a solution. Hence the two solutions : $f(x)=0$ $\forall x$ $f(x)=1$ $\forall x\ne 0$ and $f(0)=a$ where $a$ is any real.

set $x=y $ to get $2xf(x)=2xf(x)^2$ so, for $x\neq 0$ we have $ f(x)[f(x)-1]=0$ so, for $x\neq 0$ either $f(x)=0 $ or $1$. now let there is $x_{1}\neq 0, x_{2}\neq 0$ s.t. $f(x_{1})=0,f(x_{2})=1$ then set $x=x_{1},y=x_{2} $ to see $x_{1}=0$ contradiction. so, either $f(x)=0$ for all $x\neq 0$ or $f(x)=1$ for all $x\neq 0$ consider $f(x)=0$ for $x\neq 0$ then set $x\neq 0$ and $y=0$ to get$xf(0)=0$ so, $f(0)=0$ so then see $f(x)=0 $ for all x is a solution. for the other case see $f(x)=1 $ for $x\neq 0$ and $f(0)=c$ satisfy the equation. so the solutions are $f(x)=1 $ for $x\neq 0$ and $f(0)=c$ where $c\in \mathbb R$ and $f(x)=0$ for all $x\in \mathbb R$

Find all functions $f:\mathbb R\to\mathbb R$ satisfying the inequality $$xf(y)+yf(x)=(x+y)f(x)f(y)$$for all reals $x,y$. Prove that exactly two of them are continuous. I. Dimovski