3. Find the smallest distance from the point with coordinates $(10 ; 5 ; 10)$ to a point whose coordinates are positive and satisfy the inequality $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 9 \sqrt{1-(2 x+y)^{2}}$. In your answer, write the square of the found distance.
Solution. We use the Cauchy inequality $a+b+c \geq 3 \sqrt[3]{a b c}$, which holds for all positive values of $a, b, c$. Then, $x+y+z \geq 3 \sqrt[3]{x y z}, \quad \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq 3 \sqrt[3]{\frac{1}{x y z}}$. Since all parts of the inequalities are positive, we have

$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 9 \sqrt[3]{x y z \cdot \frac{1}{x y z}}=9 . \quad \text { The expression }
$$

$9 \sqrt{1-(2 x+y)^{2}} \leq 9$ for all possible values of $x$ and $y$. The original inequality is valid for all positive $z$ and all positive $x$ and $y$ for which the inequality 1 $(2 x+y)^{2} \geq 0$ holds. Thus, the minimum distance from the point with coordinates $(10 ; 5 ; 10)$ to the point, the coordinates of which are positive and satisfy the original inequality, is the distance from the point (10; 5) in the Oxy plane to the line $2 x+y=1$. This distance squared is 115.2.

Answer: 115.2.