2. Positive numbers $a, b, c$ are such that $a^{2} b+b^{2} c+c^{2} a=3$. Find the minimum value of the expression

$$
A=\frac{\sqrt{a^{6}+b^{4} c^{6}}}{b}+\frac{\sqrt{b^{6}+c^{4} a^{6}}}{c}+\frac{\sqrt{c^{6}+a^{4} b^{6}}}{a} .
$$
Answer: $3 \sqrt{2}$.

Solution. We will use the inequality $\sqrt{x+y} \geqslant \frac{1}{\sqrt{2}}(\sqrt{x}+\sqrt{y})$ for $x, y \geqslant 0$. Taking into account the Cauchy inequality, we get

$$
\begin{aligned}
& A \geqslant \frac{1}{\sqrt{2}}\left(\frac{a^{3}+b^{2} c^{3}}{b}+\frac{b^{3}+c^{2} a^{3}}{c}+\frac{c^{3}+a^{2} b^{3}}{a}\right)=\frac{1}{\sqrt{2}}\left(\frac{a^{3}}{b}+b c^{3}+\frac{b^{3}}{c}+c a^{3}+\frac{c^{3}}{a}+a b^{3}\right) \geqslant \\
& \geqslant \frac{1}{\sqrt{2}}\left(\left(\frac{a^{3}}{b}+a b^{3}\right)+\left(\frac{b^{3}}{c}+b c^{3}\right)+\left(\frac{c^{3}}{a}+c a^{3}\right)\right) \geqslant \sqrt{2}\left(a^{2} b+b^{2} c+c^{2} a\right)=3 \sqrt{2}
\end{aligned}
$$

Equality is achieved when $a=b=c=1$.