The anwser is f(x) = cx^2 (with c is a positif integer or 0)

We claim that the answer is $\boxed{f(x) = cx^2}$ with $c \geq 0$ a nonnegative real. To see that this works, apply AM-GM: \[2x^3 z^3 + y^3 = x^3 z^3 + x^3 z^3 + y^3 \geq 3x^2 yz^2.\]We now show that this is the only family of solutions. Let $P(x,y,z)$ denote the proposition $2x^2z f(z) + yf(y) \geq 3yz^2 f(x)$. $P(0,1,z): f(1) \geq 3z^2 f(0)$. Since $z$ is arbitrary, this means that $f(0)=0$. $P(x,x,1): 2x^3 f(1) + xf(x) \geq 3xf(x) \quad\Rightarrow\quad 2x^3 f(1) \geq 2xf(x)\quad\Rightarrow\quad f(x) \leq cx^2$ where $c = f(1)$. $P(1,x,x): 2x f(x) + xf(x) \geq 3cx^3 \quad\Rightarrow\quad f(x) \geq cx^2$. Combining the two inequalities, we get $f(x) = cx^2$ as desired.

Claim: $f(x)=cx^2$ for some $c\ge 0$. Proof: Let $f(x)=x^2g(x)$. Note that $2x^3z^3g(z)+y^3g(y)\ge 3yx^2z^2g(x)$. By putting $y=xy$ we get that $x^3z^3g(xz)\ge x^3z^3g(x)$. If $x,z>0$ it's clear that $g(x)=c$ for all positive reals $x$ because $g(t)\ge g(s)$ and $g(s)\ge g(t)$ for all positive reals $s,t$. Thus $f(x)=cx^2$ for all positive reals $x$. You can now get that $f(0)=0$ and hence $f(x)=cx^2$

ACGNmath wrote: $P(0,1,z): f(1) \geq 3z^2 f(0)$. Since $z$ is arbitrary, this means that $f(0)=0$. How do we conclude this? Can't $f(0)$ be negative?

Sedro wrote: ACGNmath wrote: $P(0,1,z): f(1) \geq 3z^2 f(0)$. Since $z$ is arbitrary, this means that $f(0)=0$. How do we conclude this? Can't $f(0)$ be negative? y=z=0 give f(0) non negatif