We have \begin{align*} f(xf(y))<f(xf(y))+f(yf(z))+f(zf(x))=xy+yz+zx\\ f(xf(y))\leq\lim_{z\searrow0}xy+yz+zy=xy \end{align*}Analogously $f(yf(z))\leq yz,f(zf(x))\leq zx$. We must have equality in \[ f(xf(y))+f(yf(z))+f(zf(x))=xy+yz+zx \]Thus $f(xf(y))=xy$ for any $x,y\in\mathbb{R}^+$. Plugging in $y=1, x=\frac{t}{f(1)}$ gives now \[ f(t)=ct \]Since $f(xf(y))=xy$, we have $c=1$. Thus $f\equiv x$ Obviously, $f(x)=x$ is a solution.

A bit strange solution. Hope it works. Let $P(x,y,z)$ denote assertion of given functional equation. Observe that $P(1,1,1)$ gives us: $$ 3f(f(1)) = 3 \implies f(f(1)) = 1 $$Thus there exists $a$ such that $f(a) =1$. On another hand $P(a,a,a)$ reveals that: $$ 3 =3a^2 \implies a =1 $$Now we see that $P(x,1,1)$ gives: $$ f(f(x)) + f(x) =2x \implies f^{n+2}(x) + f^{n+1}(x) = 2f^n(x) $$If we define sequence as $g_n = f^n(x) $, then it satisfies $g_{n+2} + g_{n+1} =2g_n $. It can be also defined as: $$ g_n = C_1 (t_1)^n + C_2(t_2)^n $$where $t_1, t_2 $ are roots of $t^2 +t -2 = 0$ and $C_1, C_2$ some constants. So in another words: $$ g_n = C_1(-2)^n + C_2 $$But for sufficiently large $n$ we will have $g_n <0$, which is impossible since $g_n=f^n(x) >0$ Consequently we have $C_1 = 0$ and $g_n$ is constant. This implies $f(f(x)) = f(x)$. To finish we note that: $$ f(f(x)) +f(x) =2x \implies 2f(x) =2x \implies f(x) =x $$which clearly works.

Greece 2020 TST P1 wrote: Let $R_+=(0,+\infty)$. Find all functions $f: R_+ \to R_+$ such that $f(xf(y))+f(yf(z))+f(zf(x))=xy+yz+zx$, for all $x,y,z \in R_+$. My solution during contest. Let $P(x,y,z)$ be the given FE. The proof constitutes of several Claims: Claim 1: $f(1)=1$ Proof: $P(x,x,x)$ yields $f(xf(x))=x^2$ so by taking $x \rightarrow 1$ here we have $f(f(1))=1$. Hence $f(1)^2=f(f(1)f(f(1)))=f(f(1))=1$, therefore $f(1)=1$, as claimed $\blacksquare$. Claim 2: $f(f(x))=2x-f(x)$ for all $x>0$ Proof: Consider $P(x,1,1)$ and conclude $\blacksquare$. Claim 3: $f(xf(y))=f(x)-f(y)+y-x+xy$ for all $x,y>0$. Proof: Consider $P(x,y,1)$ and the result of Claim 2 and conclude $\blacksquare$. Claim 4: $f \equiv x$ Proof: Take $x \rightarrow f(x)$ at $f(xf(y))=f(x)-f(y)+y-x+xy$ and obtain $f(f(x)f(y))=f(f(x))-f(y)+y-f(x)+f(x)y$, so by interchanging $x$ and $y$ and equating we obtain $$f(f(x))-f(y)+y-f(x)+f(x)y=f(f(y))-f(x)+x-f(y)+f(y)x$$which after using $f(f(x))=2x-f(x)$ and $f(f(y))=2y-f(y)$ transforms to $(f(x)-1)(y-1)=(f(y)-1)(x-1)$. Hence, for all $x,y \neq 1$, $$\frac{f(x)-1}{x-1}=\frac{f(y)-1}{y-1,}$$therefore $f(x)-1=c(x-1)$ for all $x \neq 1$ with $c$ constant. But, note that since $f(1)=1$, the previous relation holds for all $x>0$. Therefore, $f(x)=cx-c+1$, which after substituting at $P(x,y,z)$ implies that $c=1$, thus $f \equiv x$, which evidently satisfies $\blacksquare$.

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Since $f(xf(y))<f(xf(y))+f(yf(z))+f(zf(x))=xy+yz+zx$, taking $z$ small enough yields $f(xf(y))\le xy$ (and cyclic variations). But then equality holds by $P(x,y,z)$, so $Q(x,y):f(xf(y))=xy$. Then $Q\left(\frac x{f(1)},1\right)\Rightarrow f(x)=cx,c\in\mathbb R^+$. Testing, only $\boxed{f(x)=x}$ works.

jasperE3 wrote: Since $f(xf(y))<f(xf(y))+f(yf(z))+f(zf(x))=xy+yz+zx$, taking $z$ small enough yields $f(xf(y))\le xy$ (and cyclic variations). But then equality holds by $P(x,y,z)$, so $Q(x,y):f(xf(y))=xy$. Since $f\not\equiv0$, $\exists j:f(j)\ne0$ and $Q\left(\frac x{f(j)},j\right)\Rightarrow f(x)=cx,c\in\mathbb R$. Testing, only $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ work. Domain and codomain are $\mathbb{R}^+$.

Note that $f(xf(y))<xy+yz+zx.$ As $z\to 0,$ we have $f(xf(y))\leq xy.$ Then due to the problem we must have $f(xf(y))=xy.$ Setting $x=x/f(1)$ and $y=1$ implies $f(x)=x/f(1).$ Checking gives $f(x)\equiv x$ which clearly satisfies.

Let $P(x,y,z)$ denote the functional equation, $\sum_{cyc}f(xf(y))=\sum_{cyc}xy$ $P(1,1,1)$ yields $3f(f(1))=3\Longrightarrow f(f(1))=1$ Furthermore notice that, $f(xf(y))<\sum_{cyc}f(xf(y))=\sum_{cyc}xy\Longrightarrow f(xf(y))\le \lim_{z\to 0}\sum_{cyc}xy=xy$ $f(yf(z))<\sum_{cyc}f(xf(y))=\sum_{cyc}xy\Longrightarrow f(yf(z))\le \lim_{x\to 0}\sum_{cyc}xy=yz$ $f(zf(x))<\sum_{cyc}f(xf(y))=\sum_{cyc}xy\Longrightarrow f(zf(x))\le \lim_{y\to 0}\sum_{cyc}xy=zx$ Thus $\sum_{cyc}f(xf(y))\le\sum_{cyc}xy$, however we must have equality from $P(x,y,z)$, thus $f(xf(y))=xy, \forall x,y\in \mathbb{R^+}$ Now, let $Q(x,y)$ denote $f(xf(y))=xy$. $Q(f(1),f(1))$ yields $f(f(1)f(f(1)))=f(1)^2\Longrightarrow f(1)^2=1\Longrightarrow f(1)=1$ $Q(1,x)$ yields $f(f(x))=x$ (!) Furthermore $P(x,1,1)$ yields $f(f(x))+f(x)=2x$, however notice that, from (!), we have that $f(f(x))+f(x)=x+f(x)=2x$ which implies that $f(x)=x$ So to sum up, $\boxed{f(x)=x, \forall x \in \mathbb{R^+}}$ $\blacksquare$