# Problem 5. (3 points)

Positive numbers $x, y$, and $z$ are such that $x+y+z=5$. What is the smallest value that the quantity $x^{2}+y^{2}+2 z^{2}-x^{2} y^{2} z$ can take?
# Answer: -6

## Solution:

Rewrite the condition as $\frac{x}{2}+\frac{x}{2}+\frac{y}{2}+\frac{y}{2}+z=5$ and write the inequality for the arithmetic mean and the quadratic mean of these five numbers:

$$
1=\frac{\frac{x}{2}+\frac{x}{2}+\frac{y}{2}+\frac{y}{2}+z}{5} \leqslant \sqrt{\frac{\frac{x^{2}}{4}+\frac{x^{2}}{4}+\frac{y^{2}}{4}+\frac{y^{2}}{4}+z^{2}}{5}}=\sqrt{\frac{x^{2}+y^{2}+2 z^{2}}{10}}
$$

from which $x^{2}+y^{2}+2 z^{2} \geqslant 10$. Now write the inequality for the arithmetic mean and the geometric mean of these same numbers:

$$
1=\frac{\frac{x}{2}+\frac{x}{2}+\frac{y}{2}+\frac{y}{2}+z}{5} \geqslant \sqrt[5]{\frac{x}{2} \cdot \frac{x}{2} \cdot \frac{y}{2} \cdot \frac{y}{2} \cdot z}
$$

from which $x^{2} y^{2} z \leqslant 16$. Therefore, $x^{2}+y^{2}+2 z^{2}-x^{2} y^{2} z \geqslant 10-16=-6$. The minimum is achieved when all numbers for which the inequalities of means are applied are equal, that is, $x=y=2$, $z=1$.

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