2. For $x, y \in[1,3]$ find the minimum value of the expression

$$
A=\frac{\left(3 x y+x^{2}\right) \sqrt{3 x y+x-3 y}+\left(3 x y+y^{2}\right) \sqrt{3 x y+y-3 x}}{x^{2} y+y^{2} x}
$$
Answer: 4.

Solution. Note that

$$
3 x y+x-3 y-x^{2}=(3 y-x)(x-1) \geqslant 0
$$

since $x \geqslant 1$ and $x \leqslant 3 \leqslant 3 y$. Then

$$
\left(3 x y+x^{2}\right) \sqrt{3 x y+x-3 y} \geqslant 3 x^{2} y+x^{3} \quad \text { and, similarly, } \quad\left(3 x y+y^{2}\right) \sqrt{3 x y+y-3 x} \geqslant 3 y^{2} x+y^{3} .
$$

Moreover, by the Cauchy inequality

$$
x^{2} y+y^{2} x=x y(x+y) \leqslant \frac{1}{4}(x+y)^{3}
$$

Therefore,

$$
A \geqslant \frac{3 x y(x+y)+x^{3}+y^{3}}{\frac{1}{4}(x+y)^{3}}=4 \cdot \frac{(x+y)^{3}}{(x+y)^{3}}=4
$$

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