3. Find the smallest distance from the point with coordinates $(5 ; 10 ; 13)$ to a point whose coordinates are positive and satisfy the inequality

$$
(x+y+z)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right) \geq 4.5 \sqrt[4]{1-(2 y+x)^{2}}
$$

In your answer, write the square of the found distance.

(points)
Solution. Using the Cauchy inequality $a+b+c \geq 3 \sqrt[3]{a b c}$, which holds for all positive values of $a, b, c$. Then $((x+y)+(y+z)+(z+x)) \geq 3 \sqrt[3]{(x+y)(y+z)(z+x)}$, $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \geq 3 \sqrt[3]{\frac{1}{(x+y)(y+z)(z+x)}}$. Since all parts of the inequalities are positive, then $(2 x+2 y+2 z)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right) \geq 9 \sqrt[3]{(x+y)(y+z)(z+x) \cdot \frac{1}{(x+y)(y+x)(z+x)}}=9$. The expression $4.5 \sqrt{1-(2 y+x)^{2}} \leq 4.5$ 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 $1-(2 y+x)^{2} \geq 0$ holds. Thus, the minimum distance from the point with coordinates $(5 ; 10 ; 13)$ to the point, the coordinates of which are positive and satisfy the original inequality, is the distance from the point $(5 ; 10)$ in the $O x y$ plane to the line $2 y+x=1$. This distance squared is 115.2.

Answer: 115.2.