2. Given $x, y, z \geqslant 3$ find the minimum value of the expression

$$
A=\frac{\left(x^{3}-24\right) \sqrt[3]{x+24}+\left(y^{3}-24\right) \sqrt[3]{y+24}+\left(z^{3}-24\right) \sqrt[3]{z+24}}{x y+y z+z x}
$$
Answer: 1.

Solution. For $x \geqslant 3$, the inequalities $\sqrt[3]{x+24} \geqslant 3$ and $x^{3} \geqslant 3 x^{2}$ hold, from which

$$
\left(x^{3}-24\right) \sqrt[3]{x+24} \geqslant 3\left(3 x^{2}-24\right)
$$

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

$$
A \geqslant 3 \cdot \frac{3 x^{2}-24+3 y^{2}-24+3 z^{2}-24}{x y+y z+z x}=9\left(\frac{x^{2}+y^{2}+z^{2}}{x y+y z+z x}-\frac{24}{x y+y z+z x}\right) \geqslant 9\left(1-\frac{24}{27}\right)=1
$$

Equality is achieved when $x=y=z=3$.