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

$$
A=\frac{a^{4}+b^{4}}{c^{2}+4 a b}+\frac{b^{4}+c^{4}}{a^{2}+4 b c}+\frac{c^{4}+a^{4}}{b^{2}+4 c a}
$$
Answer: $\frac{6}{5}$.

Solution. By the inequalities for means

$$
\frac{a^{4}+b^{4}}{c^{2}+4 a b} \geqslant \frac{a^{4}+b^{4}}{c^{2}+2 a^{2}+2 b^{2}}=\frac{a^{4}+b^{4}}{3+a^{2}+b^{2}} \geqslant \frac{\left(a^{2}+b^{2}\right)^{2}}{2\left(3+a^{2}+b^{2}\right)}=\frac{\left(3-c^{2}\right)^{2}}{2\left(6-c^{2}\right)}=\frac{1}{2}\left(-c^{2}+\frac{9}{6-c^{2}}\right)
$$

Similarly, the other two terms in $A$ are estimated. Adding these inequalities, we get

$$
A \geqslant \frac{1}{2}\left(-\left(a^{2}+b^{2}+c^{2}\right)+\frac{9}{6-a^{2}}+\frac{9}{6-b^{2}}+\frac{9}{6-c^{2}}\right) \geqslant \frac{1}{2}\left(-3+\frac{81}{18-\left(a^{2}+b^{2}+c^{2}\right)}\right)=\frac{6}{5}
$$

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