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

$$
A=\frac{(x+2 y) \sqrt{x+y-x y}+(y+2 z) \sqrt{y+z-y z}+(z+2 x) \sqrt{z+x-z x}}{x y+y z+z x}
$$
Answer: 3.

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

$$
\sqrt{x+y-x y}=\sqrt{x+y(1-x)} \geqslant \sqrt{x} \geqslant x
$$

Then $(x+2 y) \sqrt{x+y-x y} \geqslant x^{2}+2 x y$ and similarly,

$$
(y+2 z) \sqrt{y+z-y z} \geqslant y^{2}+2 y z, \quad(z+2 x) \sqrt{z+x-z x} \geqslant z^{2}+2 z x
$$

Therefore,

$$
A \geqslant \frac{x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x}{x y+y z+z x}=\frac{(x+y+z)^{2}}{x y+y z+z x} \geqslant 3
$$

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