2. For $x, y, z \in (0,2]$, find the maximum value of the expression

$$
A=\frac{\left(x^{3}-6\right) \sqrt[3]{x+6}+\left(y^{3}-6\right) \sqrt[3]{y+6}+\left(z^{3}-6\right) \sqrt[3]{z+6}}{x^{2}+y^{2}+z^{2}}
$$
Answer: 1.

Solution. For $x \in(0,2]$, the inequalities $\sqrt[3]{x+6} \leqslant 2$ and $x^{3} \leqslant 2 x^{2}$ hold, from which

$$
\left(x^{3}-6\right) \sqrt[3]{x+6} \leqslant 2\left(2 x^{2}-6\right)
$$

Similarly, the other two terms in the numerator of $A$ are estimated. Therefore,

$$
A \leqslant 2 \cdot \frac{2 x^{2}-6+2 y^{2}-6+2 z^{2}-6}{x^{2}+y^{2}+z^{2}}=4-\frac{36}{x^{2}+y^{2}+z^{2}} \leqslant 4-\frac{36}{12}=1
$$

Equality is achieved when $x=y=z=2$.