3. Find the smallest distance from the point with coordinates $(7 ; 3 ; 6)$ to a point whose coordinates are positive and satisfy the inequality

$$
\left(x^{2}+y^{2}+z^{2}\right)\left(\frac{1}{x y}+\frac{1}{y z}+\frac{1}{x z}\right) \geq 9 \sqrt{1-(2 z+y)^{2}}
$$

In your answer, write the square of the found distance.
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 $x^{2}+y^{2}+z^{2} \geq 3 \sqrt[3]{x^{2} y^{2} z^{2}}$, $\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \geq 3 \sqrt[3]{\frac{1}{x^{2} y^{2} z^{2}}}$. Since all parts of the inequalities are positive, then $\left(x^{2}+y^{2}+z^{2}\right)\left(\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}\right) \geq 9 \sqrt[3]{x^{2} y^{2} z^{2} \cdot \frac{1}{x^{2} y^{2} z^{2}}}=9$. The expression $9 \sqrt{1-(2 z+y)^{2}} \leq 9$ for all possible values of $z$ and $y$. The original inequality is valid for all positive $x$ and all positive $z$ and $y$ for which the inequality $1-(2 z+y)^{2} \geq 0$ holds. Thus, the minimum distance from the point with coordinates $(7 ; 3 ; 6)$ to the point, the coordinates of which are positive and satisfy the original inequality, is the distance from the point $(3 ; 6)$ in the $O y z$ plane to the line $2 z+y=1$. This distance squared is 39.2.

Answer: $39.2$