2. Given $a, b, c > 0$, find the maximum value of the expression

$$
A=\frac{a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)}{a^{3}+b^{3}+c^{3}-2 a b c}
$$
Answer: 6.

Solution. By the Cauchy-Schwarz inequality,

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

Note that

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

Let $t=\frac{(a+b+c)^{3}}{a b c}$. By the Cauchy-Schwarz inequality, $t \geqslant 27$, hence

$$
A \leqslant \frac{3\left((a+b+c)^{3}-9 a b c\right)}{(a+b+c)^{3}-18 a b c}=\frac{3(t-9)}{t-18}=3\left(1+\frac{9}{t-18}\right) \leqslant 6
$$

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