2. Given $x, y, z \geqslant 1$, find the minimum value of the expression

$$
A=\frac{\sqrt{3 x^{4}+y}+\sqrt{3 y^{4}+z}+\sqrt{3 z^{4}+x}-3}{x y+y z+z x}
$$
Answer: 1.

Solution. Note that for $x, y \geqslant 1$

$$
\sqrt{3 x^{4}+y}-1=\sqrt{x^{4}+2 x^{4}+y}-1 \geqslant \sqrt{x^{4}+2 x^{2}+1}-1=x^{2}
$$

Similarly, it can be shown that

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

Therefore,

$$
A \geqslant \frac{x^{2}+y^{2}+z^{2}}{x y+y z+z x} \geqslant 1
$$

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