IMO Team Selection Test 1

$$f_{c,d}(x)=\{\begin{array}{ll}c{x}^2& \text{if }x>0\\ d& \text{if }x=0\end{array}$$ with $c\ge 0$ and $d\le 0$.

Substituting $x=0$ and $y=1$ yields $f(1)\ge 3f(0)z^2$ for all $z\ge 0$. If $f(0)>0$, then the right-hand side of this inequality is unbounded, contradiction. So we have $f(0)\le 0$.

Substituting $z=0$ yields $yf(y)\ge 0$, so $f(y)\ge 0$ for all $y>0$. In particular, $f(1)\ge 0$. Substituting $x=y$ and $z=1$ yields $$2y^{3}f(1)+yf(y)\ge 3yf(y),$$ so $2y^{3}f(1)\ge 2yf(y)$, so $f(y)\le y^{2}f(1)$ for $y>0$.

Substituting $z=y$ and $x=1$ yields $$2yf(y)+yf(y)\ge 3y^{3}f(1),$$ so $3yf(y)\ge 3y^{3}f(1)$, so $f(y)\ge y^{2}f(1)$ for $y>0$.

Together, this yields $f(y)=y^{2}f(1)$ for $y>0$. Write $c=f(1)$ and $d=f(0)$. So then we now know that $$f(x)=\{\begin{array}{ll}c{x}^2& \text{if }x>0\\ d& \text{if }x=0\end{array}$$ with $c\ge 0$ and $d\le 0$. We will now check these functions.

Note first that $xf(x)\ge 0$ for all $x\in {\mathbb{R}}_{\ge 0}$: for $x=0$ this is trivial and for $x>0$ $xf(x)=c{x}^3\ge 0$ since $c\ge 0$.

For $x=0$ the inequality becomes $yf(y)\ge 3y{z}^{2}f(0)$, and this is correct since $yf(y)\ge 0$ and $3y{z}^{2}f(0)=3y{z}^{2}d\le 0$ for $y,z\ge 0$.

For $y=0$ the inequality becomes $2x^{3}zf(z)\ge 0$ and this is correct. For $z=0$ the inequality becomes $yf(y)\ge 0$ and this is correct.

Now assume that $x,y,z>0$. So then $f(x)=c{x}^2$, $f(y)=c{y}^2$, $f(z)=c{z}^2$ and we have to prove that $$2x^{3}zc{z}^2+yc{y}^2\ge 3y{z}^{2}c{x}^2.$$ Since $c\ge 0$, it is sufficient to show that $2x^3{z}^3+y^3\ge 3y{z}^2{x}^2$ for all $x,y,z>0$. This follows from the inequality of the arithmetic and geometric mean on the three terms $x^3{z}^3$, $x^3{z}^3$ and $y^3$. So all these functions satisfy the requirements. $\square$