10. find all pairs $(\alpha, \beta)$ of positive real numbers with the following properties:

(a) For all positive real numbers $x, y, z, w$ applies

$$
x+y^{2}+z^{3}+w^{6} \geq \alpha(x y z w)^{\beta}
$$

(b) There is a quadruple $(x, y, z, w)$ of positive real numbers such that equality holds in (a).

## Solution
Let $(\alpha, \beta)$ be a pair as in the problem. We first note that for fixed $\beta$ there exists at most one $\alpha$ as in (b), because $\alpha$ is then the largest positive real number for which (a) applies. Now set $x=t^{6}, y=t^{3}, z=t^{2}$ and $w=t$ for $t>0$, then in particular $4 t^{6} \geq \alpha t^{12 \beta}$, i.e.

$$
t^{6-12 \beta} \geq \frac{\alpha}{4}
$$

for all $t>0$. For very small $t$, this estimate can only apply if the exponent on the left-hand side is not negative. For large $t$, on the other hand, only if the exponent is not positive. Consequently, $\beta=\frac{1}{2}$ must be.

On the other hand, AM-GM

$$
\begin{aligned}
x+y^{2}+z^{3}+w^{6} & =(\underbrace{\frac{x}{6}+\ldots+\frac{x}{6}}_{6})+\left(\frac{y^{2}}{3}+\frac{y^{2}}{3}+\frac{y^{2}}{3}\right)+\left(\frac{z^{3}}{2}+\frac{z^{3}}{2}\right)+w^{6} \\
& \geq 12 \sqrt[12]{\left(\frac{x}{6}\right)^{6} \cdot\left(\frac{y^{2}}{3}\right)^{3} \cdot\left(\frac{z^{3}}{2}\right)^{2} \cdot w^{6}}=\left(2^{\frac{4}{3}} 3^{\frac{1}{4}}\right) \sqrt{x y z w} .
\end{aligned}
$$

Equality applies to $(x, y, z, w)=(6, \sqrt{3}, \sqrt[3]{2}, 1)$, so (a) and (b) are fulfilled for the only pair $(\alpha, \beta)=\left(2^{\frac{4}{3}} 3^{\frac{1}{4}}, \frac{1}{2}\right)$.