Let $\alpha$ be a positive real number. Determine for this $\alpha$ the largest real number $C$ such that for all positive real numbers $x, y,$ and $z$ with $xy + yz + zx = \alpha$, the inequality

$$
\left(1+\frac{\alpha}{x^{2}}\right)\left(1+\frac{\alpha}{y^{2}}\right)\left(1+\frac{\alpha}{z^{2}}\right) \geq C \cdot\left(\frac{x}{z}+\frac{z}{x}+2\right)
$$

holds. When does equality hold?

(Walther Janous)
We substitute the side condition for $\alpha$, multiply both sides by the common denominator $x^{2} y^{2} z^{2}>0$, and obtain the equivalent inequality

$$
\left(x^{2}+x y+x z+y z\right)\left(y^{2}+y x+y z+x z\right)\left(z^{2}+z x+z y+x y\right) \geq C x y^{2} z\left(x^{2}+z^{2}+2 x z\right) .
$$

The inequality is now homogeneous of degree 6, so the side condition is no longer relevant. Each of the factors can be factored:

$$
(x+y)(x+z)(y+x)(y+z)(z+x)(z+y) \geq C^{2} z(x+z)^{2} .
$$

We cancel $(x+z)^{2}>0$ and obtain the equivalent inequality

$$
(x+y)^{2}(z+y)^{2} \geq C x y^{2} z .
$$

If we estimate each of the two factors on the left using the arithmetic-geometric mean inequality, we obtain an optimal value of $C=16$, which is achieved for $x=y=z=\sqrt{\alpha / 3}$.

(Clemens Heuberger)