Problem 1756

$$f_{c,d}(x) = \begin{cases} cx^2 & \text{if } x > 0 \\ d & \text{if } x = 0 \end{cases}$$
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
$$2y^3f(1) + yf(y) \ge 3yf(y),$$
so $2y^3f(1) \ge 2yf(y)$, so $f(y) \le y^2f(1)$ for $y > 0$.

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

Together, this yields $f(y) = y^2f(1)$ for $y > 0$.
Write $c = f(1)$ and $d = f(0)$. So then we now know that
$$f(x) = \begin{cases} cx^2 & \text{if } x > 0 \\ d & \text{if } x = 0 \end{cases}$$
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 $x > 0$ $xf(x) = cx^3 \ge 0$ since $c \ge 0$.

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

For $y = 0$ the inequality becomes $2x^3zf(z) \ge 0$ 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 $f(x) = cx^2$, $f(y) = cy^2$, $f(z) = cz^2$ and we have to prove that
$$2x^3zcz^2 + ycy^2 \ge 3yz^2cx^2.$$
Since $c \ge 0$, it is sufficient to show that $2x^3z^3 + y^3 \ge 3yz^2x^2$ for all $x, y, z > 0$. This follows from the inequality of the arithmetic and geometric mean on the three terms $x^3z^3$, $x^3z^3$ and $y^3$. So all these functions satisfy the requirements. $\square$