WRONG SOLUTION sorry Setting $x=y=z=0$ we have $f(0)=0$ Now put $z=0$. Thus $f(xf(y))=yf(x).....(i)$ Let $f(1)=a$ putting $y=1$ in $(i)$ we have $f(ax)=f(x)$ Again, interchanging positions of $x$ and $a$, $f(ax)=xf(1)=ax$ thus comparing we have, $f(x)=ax$ which is indeed a solution for any $a\in\mathbb R$

Abhinandan18 wrote: Again, interchanging positions of $x$ and $a$, $f(ax)=xf(1)=ax$ You can't do that. $a=f(1)$ is fixed.

Sorry, ignore this post.

Are you sure that $P(f(x),f(y),f(z))$ gives that? This is not an easy FE!

It does not give that

Easy that $f(0)=0$, $f(1)=1$, $f(f(x))=x$ and $f(xy)=f(x)f(y)$. From $f(xy)=f(x)f(y)$, we get $f(x)\geq 0$ for all $x\geq 0$. Also, since $f$ is bijective, $f(-1)=-1$. Denote that, if $2n-1$ is fixed point ($x$ that satisfies $f(x)=x$), then $\sqrt{2n-1}$ and $\frac{1}{\sqrt{2n-1}}$ also is. Then $P \left(\sqrt{2n-1},\sqrt{2n-1},\frac{1}{\sqrt{2n-1}} \right)$ gives us $f(2n+1)=2n+1$. Therefore we can prove that $f(2n+1)=2n+1$, for all nonnegative integer $n$. Since $f(-x)=-f(x)$ and $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$, we get all elements of set $Q_{odd}=\{\frac{2m+1}{2n+1}$ where $x,y \in \mathbb{Z}\}$ are fixed point. Also, we know any open interval $(a,b)$ contains element of $Q_{odd}$. Now to prove $f(z)=z$ for all $z$, suppose not. Our goal is to find $x,y \in Q_{odd}$ such that $xy+yf(z)+zx>0$ and $xy+yz+xf(z)<0$, which will contradict to $f(x)\geq 0$ for all $x\geq 0$. Case 1: $f(z)<z$ Easy. Set $y\in Q_{odd}$ such that $y+z>0$ and $y+f(z)<0$, and $x\in Q_{odd}$ such that \[x>\max \{ \frac{-yf(z)}{y+z},\frac{-yz}{y+f(z)} \}\]so we're done. Case 2: $f(z)>z$ Similar. Set $y\in Q_{odd}$ such that $y+z<0$ and $y+f(z)>0$, and $x\in Q_{odd}$ such that \[x<\min \{ \frac{-yf(z)}{y+z},\frac{-yz}{y+f(z)} \}\]then we're done. Finally, we get $f(x)=x$ for all $x$, as desired. Interesting.. Is there more solution?

IsoLyS wrote: Easy that $f(0)=0$, $f(1)=1$, $f(f(x))=x$ and $f(xy)=f(x)f(y)$. From $f(xy)=f(x)f(y)$, we get $f(x)\geq 0$ for all $x\geq 0$. Also, since $f$ is bijective, $f(-1)=-1$. Denote that, if $2n-1$ is fixed point ($x$ that satisfies $f(x)=x$), then $\sqrt{2n-1}$ and $\frac{1}{\sqrt{2n-1}}$ also is. Then $P \left(\sqrt{2n-1},\sqrt{2n-1},\frac{1}{\sqrt{2n-1}} \right)$ gives us $f(2n+1)=2n+1$. Therefore we can prove that $f(2n+1)=2n+1$, for all nonnegative integer $n$. Since $f(-x)=-f(x)$ and $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$, we get all elements of set $Q_{odd}=\{\frac{2m+1}{2n+1}$ where $x,y \in \mathbb{Z}\}$ are fixed point. Also, we know any open interval $(a,b)$ contains element of $Q_{odd}$. Now to prove $f(z)=z$ for all $z$, suppose not. Our goal is to find $x,y \in Q_{odd}$ such that $xy+yf(z)+zx>0$ and $xy+yz+xf(z)<0$, which will contradict to $f(x)\geq 0$ for all $x\geq 0$. Case 1: $f(z)<z$ Easy. Set $y\in Q_{odd}$ such that $y+z>0$ and $y+f(z)<0$, and $x\in Q_{odd}$ such that \[x>\max \{ \frac{-yf(z)}{y+z},\frac{-yz}{y+f(z)} \}\]so we're done. Case 2: $f(z)>z$ Similar. Set $y\in Q_{odd}$ such that $y+z<0$ and $y+f(z)>0$, and $x\in Q_{odd}$ such that \[x<\min \{ \frac{-yf(z)}{y+z},\frac{-yz}{y+f(z)} \}\]then we're done. Finally, we get $f(x)=x$ for all $x$, as desired. Interesting.. Is there more solution? You've missed the trivial $f(x)=0$ case.

If $f(1)=0$, $f(x)=0$: post #5 If $f(1) \neq 0$, $f(f(x))=x$ and $f(xy)=f(x)f(y)$, hence $f(1)=1$ and continues. I was asking for other solving strategies. Sorry about it. Also don't quote whole text.

IsoLyS wrote: Then $P \left(\sqrt{2n-1},\sqrt{2n-1},\frac{1}{\sqrt{2n-1}} \right)$ gives us $f(2n+1)=2n+1$. Therefore we can prove that $f(2n+1)=2n+1$, for all nonnegative integer $n$. [...] Now to prove $f(z)=z$ for all $z$, suppose not. Our goal is to find $x,y \in Q_{odd}$ such that $xy+yf(z)+zx>0$ and $xy+yz+xf(z)<0$, which will contradict to $f(x)\geq 0$ for all $x\geq 0$. I was just wondering, what is the motivation for wanting to prove that $f(2n-1)=2n-1$ implies $f(2n+1)=2n+1$ and wanting to find $x,y \in Q_{odd}$ such that $xy+yf(z)+zx>0$ and $xy+yz+xf(z)<0$ for all $z$? Thank you so much for all your help!

Korea Winter Test 2015-16 wrote: Determine all the functions $f : \mathbb{R}\rightarrow\mathbb{R}$ that satisfy the following $$f(xf(y)+yf(z)+zf(x))=yf(x)+zf(y)+xf(z)$$for all $x, y, z \in \mathbb{R}$. Nice problem!

@MathPanda1, my proof here uses a different way to reach the same conclusion, if that's of any use to you

Here is a solution by my friend and teammate RAMİL JABİYEV. İ will not give a proof for small things like multipilcavity,involutivity and odd of $f $. We try to find $k $ such that $f (\frac {k}{f (y)})+y $ is surjective. We have $f (\frac {k}{f (y)})+y=\frac {f (k)}{y}+y=a $. $\implies $ $y^2-ay+f (k)=0$ and if we let $f (k) $ to be negative we are done .By multiplicavity $f $ is negative in negatives and vice versa. So by selecting $k $ negative we are done. Then $P (\frac {-x^2}{f (y)},y,1)+P (\frac {-x^2}{f (y)},y,-1) $ and using what we have gotten so far $f (x^2+y)+f (x^2-y)=2f (x^2) $ So for any $x,y $ such that their sum is postive we have $f (x)+f (y)=2f (\frac {x+y}{2}) $. So using the fact that $f $ is odd we conclude that the above relation is true for any reals $x,y $. It is easy to show that $f $ is additive from here. So we have two solutions 1)$f\equiv 0$ 2) $f (x)=x $.

here is my solution i hope it is correct. the solutions are $f(x)=x,f(x)=0$. let $P(x,y,z)$ denote the equation and consider that $f \neq 0$. \[P(0,0,0) \implies f(0)=0 \]\[P(x,x,x) \implies f(3f(x))=3f(x) \quad \text{now we need to prove that the function is injective}\] \[P(x,0,z) \implies f(zf(x))=xf(z) \quad (I)\] in order to prove injectivity we consider that $f(x_1)=f(x_2)$ then putting $x_1$ and $x_2$ instead of $x$ in $(I)$ then comparing them we get \[x_1f(z)=x_2f(z)\implies x_1=x_2\quad \text{so the function is injective} \]so we are done.

iman007 wrote: \[P(x,x,x) \implies f(3f(x))=3f(x) \quad \text{now we need to prove that the function is injective}\] Uhh ? What injectivity would bring here ?

Another nice way to finish after finding $f(f(x))=x$, $f(xy)=f(x)f(y)$ and $f(1)=1$: Taking $x=y=-1$ in the multiplicativity formula gives $1=f(-1)^2 \implies f(-1)=-1$ (it can't be $1$ due to bijectivity). Taking $x=-1$ then gives $f(-x)=-f(x)$, and finally taking $y = -x$ gives $f(-x^2)=-f(x)^2\leq 0 \implies$ $f(x)$ and $x$ are of the same sign. $P(x, 1, -1)\implies f(x-1-f(x))=f(x)-x-1$, so we must have $x-1-f(x)$ and $f(x)-x-1$ of the same sign $\implies x-1\leq f(x) \leq x+1$ Multiplicativity easily implies $f(x^n) = f(x)^n$ for positive integers $n$ $\implies x^n-1 \leq f(x)^n \leq x^n+1 \implies (x^n-1)^{1/n} \leq f(x) \leq (x^n+1)^{1/n}$. Take positive $x$ for simplicity, and let $n \to \infty$. This easily implies $f(x)=x$ for all positive, and thus negative $x$ $\square$

johnkwon0328 wrote: Determine all the functions $f : \mathbb{R}\rightarrow\mathbb{R}$ that satisfies the following. $f(xf(y)+yf(z)+zf(x))=yf(x)+zf(y)+xf(z)$ Hard..... Case 1:$f(x)=0,x\in R$ Case 2:$f(a)\neq 0$ I solved this case below: Claim 1.$f(0)=0$ Proof: $P(0,0,0)\implies f(f(0))=0$ Claim 2.$f$ is injective function Proof: $f(a)=f(b)$ $P(x,a,0)\implies f(xf(a))=af(x)$ $P(x,b,0)\implies f(xf(b))=bf(x)$ $af(x)=f(xf(a))=f(xf(b))=bf(x)\implies a=b$ Note:$f(x)\neq 0$.(Case 2.) Claim 3.$f(\frac{1}{3})=\frac{1}{3}$ Proof: $P(\frac{1}{3},\frac{1}{3},\frac{1}{3})\implies f(f(\frac{1}{3}))=f(\frac{1}{3})\implies f(\frac{1}{3})=\frac{1}{3}$ Claim 4.$f(1)=1$ Proof: $P(1,\frac{1}{3},0)\implies f(1)=1$ Claim 5.$f(f(x))=x$ Proof: $P(1,x,0)\implies f(f(x))=x$ Claim 6.$f(xy)=f(x)f(y)$ Proof: $P(x,f(y),0)\implies f(xy)=f(x)f(y)$ Claim 7.$f(-x)=-f(x)$ Proof: $1=f(1)=f((-1)\cdot (-1))=f(-1)\cdot f(-1)=f(-1)^2$ $f(-1)=1\implies f(-1)=f(1)\implies -1\neq 1-impossible$ Then $f(-1)=-1$ $f(-x)=f(-1)f(x)=-f(x)$ Claim 8.$f(x+1)=f(x)+1$ Proof: $P(\frac{-1}{f(x)},x,1)\implies f(-1+x-\frac{1}{x})=-1+f(x)-\frac{1}{f(x)}$ $P(\frac{-1}{f(x)},x,-1)\implies f(-1-x+\frac{1}{x})=-1-f(x)+\frac{1}{f(x)}$ Then $f(-1+x-\frac{1}{x})+f(-1-x+\frac{1}{x})=-2,x-\frac{1}{x}=a$ Then $f(a+1)=f(a-1)+2$ $a-1=t\implies f(t+2)=f(t)+2\implies f(2)=2$ $t=2x\implies f(2x+2)=f(2x)+2\implies f(x+1)=f(x)+1$ Finish: $x\to \frac{x}{y}; x,y>0$ Then $f(\frac{x}{y}+1)=f(\frac{x}{y})+1\implies f(y)f(\frac{x}{y}+1)=f(y)(f(\frac{x}{y})+1)$ Then $f(x+y)=f(x)+f(y)-additive$ And $f(xy)=f(x)f(y)$ Then $f(x)=x$

Let $P(x,y,z)$ denote the original expression. Clearly $$\boxed{f(x)=0 \ \ \forall x \in \mathbb{R}}$$So assume $f$ is not all-zero from now on. $P(0,0,0)$ $\implies$ $f(0)=0$. $P(x,y,0)$ $\implies$ $f(xf(y))=yf(x)$ $$\implies \ f(f(xf(y)))=f(yf(x))=xf(y)$$Since $f$ is not all-zero, we get $f(f(z))=z$ for all $z$. Therefore $f$ is bijective. Now, $P(1,y,0)$ for some $y \neq 0$ gives $f(1)=1$. Also, $P(x,f(y),0)$ $\implies$ $f(xy)=f(x)f(y)$ for all $x,y$. In particular, $f(x)>0$ for $x>0$. Let $x=-f(z)$ for an arbitrary $z$. Then $P(x,1,z)$ gives $$f(zf(x))=f(x)+z+xf(z)$$$$\implies \ f(-f(z))=-z$$for all $z$. Putting $t=f(z)$ in the above we get $f(-t)=-f(t)$ for all $t$. So comparing $P(x,y,z)$ and $P(x,y,-z)$ for $x,y,z>0$ gives \begin{align} f(a+b)+f(a-b)=2f(a) \end{align}whenever $a=xf(y)$ and $b=yf(z)+zf(x)$; equivalently, $b= \dfrac{f(a)}{t}f(z)+zt$ where $t=f(x)$. Since $t$ can be arbitrary (but positive), $(1)$ holds whenever $b \geq 2f(a) \sqrt{zf(z)}$ for some $z>0$ by AM-GM. Define $g(x)=xf(x)>0$ for $x>0$. Clearly $g(x)=1$ cannot hold for all $x$. Also, $g$ is multiplicative. Therefore, $g(z)$ can come as close to $0$ as possible. So, $(1)$ holds for all $a,b>0$. Call it $Q(a,b)$. $Q(a,a)$ $\implies$ $f(2a)=2f(a)$ $\implies$ $f(a+b)+f(a-b)=f(2a)$ for all $a,b>0$. This, combined with oddness, gives us that $f$ is additive. It is well known that the only function which is both additive and multiplicative over $\mathbb{R}$ is $$\boxed{f(x)=x \ \ \forall x \in \mathbb{R}}$$ $\blacksquare$

Obviously, $f(x)=0$ for all $x$ works. Otherwise, let $t$ be a real number such that $f(t)$ is nonzero; say that it equals $T$. We'll show that the only other solution is $f(x)\equiv x$. Plugging $x=y=z=0$, we get $f(0)=0$. Next, plugging $x=0$, we get \[f(yf(z))=zf(y).\]Next, swapping $y$ and $z$ gives \[f(zf(y))=yf(z).\]Hence, $f(f(yf(z)))=yf(z)$. In particular, $z=t$ gives \[f(f(yT)) = yT\]for all $y$; varying $y$ gives that $f$ is a bijection on the real numbers. (Do I use this later? I don't think so, but I might have subconsciously so I'm keeping this in here just in case ) Also, replacing $z$ with $f(z)$ and then using the bijective property gives us that $f(y)f(z)=f(yz)$, so $f$ is multiplicative. Finally, let $a$, $b$, and $c$ be nonzero real numbers, and consider setting $x=\frac{a}{f(b)}$, $y=\frac{b}{f(c)}$, $z=\frac{c}{f(a)}$. This gives that \[f\left(\frac ac+\frac cb+\frac ba\right) = f\left(\frac ac\right)+f\left(\frac cb\right)+f\left(\frac ba\right)\] Since for any real numbers $p$, $q$, $r$ with $pqr=1$ we can find $a$, $b$, $c$ such that $p=\frac ac$, $q=\frac cb$, $r=\frac ba$, for \emph{any} real numbers $p$, $q$, $r$ with product $1$, we have \[f(p+q+r)=f(p)+f(q)+f(r).\]Also, notice that if real numbers $u$, $v$, $w$ have nonzero product $P$, then we have \begin{align*} f(u+v+w)=f\left(\sqrt[3]{P}\right)\cdot f\left(\frac{u}{\sqrt[3] P}+\frac{v}{\sqrt[3] P}+\frac{w}{\sqrt[3] P}\right) &= f\left(\sqrt[3]{P}\right)\cdot f\left(\frac{u}{\sqrt[3] P}+\frac{v}{\sqrt[3] P}+\frac{w}{\sqrt[3] P}\right)\\ &=f\left(\sqrt[3]{P}\right)\cdot \left(f\left(\frac{u}{\sqrt[3] P}\right)+\left(\frac{v}{\sqrt[3] P}\right)+\left(\frac{w}{\sqrt[3] P}\right)\right)\\ &= f(u)+f(v)+f(w). \end{align*} Next, we compute $f(2)$: We have that $f(3)=f(1+1+1)=3f(1)=3\implies $ $3=f\left(\frac12 +\frac12 + 2\right)=2f\left(\frac12\right)+f(2)=\frac{2}{f(2)}+f(2)$ $\stackrel{f(2)\neq 0}{\implies} f(2)^2-3f(2)+2=0\implies f(2)=1 \text{ or } f(2)=2$. By bijectivity, $f(2)=2$. Now, taking $u=v$ in our ``almost additive" identity, we have that \[f(2u+w)=2f(u)+f(w)=f(2)f(u)+f(w)=f(2u)+f(w)\]for any nonzero $u$ and $w$. Obviously, it also holds if either $u$ or $w$ is zero, so we know that $f$ is additive. The only additive and multiplicative functions are $f(x)\equiv 0$ and $f(x)\equiv x$, so we force $f(x)=x$ for all $x$ and we're done.

Redacted

@MathLuis, isn't showing the multplicativity as well as additivity structure needed, to conclude the only solutions are f(X) = 0 or X?

MathLuis, your proof is wrong. Firstly, when you first cite $P(1,-1,x)$ and $P(1,1,x)$, you calculate it incorrectly. Secondly, there exist functions which aren't odd or even.