2. For $x, y, z \in(0,1]$ find the maximum value of the expression

$$
A=\frac{\sqrt{8 x^{4}+y}+\sqrt{8 y^{4}+z}+\sqrt{8 z^{4}+x}-3}{x+y+z}
$$
Answer: 2.

Solution. Note that for $x, y \in(0,1]$

$$
\sqrt{8 x^{4}+y}-1=\sqrt{4 x^{4}+4 x^{4}+y}-1 \leqslant \sqrt{4 x^{2}+4 x+1}-1=2 x .
$$

Similarly, it can be shown that

$$
\sqrt{3 y^{4}+z}-1 \leqslant 2 y, \quad \sqrt{3 z^{4}+x}-1 \leqslant 2 z
$$

Therefore,

$$
A \leqslant \frac{2 x+2 y+2 z}{x+y+z}=2
$$

Equality is achieved when $x=y=z=1$.