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

$$
A=\frac{a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b)}{(a+b+c)^{4}-79(a b c)^{4 / 3}}
$$
Answer: 3.

Solution. By the Cauchy-Schwarz inequality,

$a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b)=(a+b+c)\left(a^{3}+b^{3}+c^{3}\right)-\left(a^{4}+b^{4}+c^{4}\right) \leqslant(a+b+c)\left(a^{3}+b^{3}+c^{3}\right)-3(a b c)^{4 / 3}$.

Note that $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6 a b c+B$, where

$$
B=3\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right) \geqslant 18 \sqrt[6]{a^{6} b^{6} c^{6}}=18 a b c
$$

(we used the Cauchy-Schwarz inequality). Then $a^{3}+b^{3}+c^{3} \leqslant(a+b+c)^{3}-24 a b c$ and

$(a+b+c)\left(a^{3}+b^{3}+c^{3}\right)-3(a b c)^{4 / 3} \leqslant(a+b+c)^{4}-24 a b c(a+b+c)-3(a b c)^{4 / 3} \leqslant(a+b+c)^{4}-75(a b c)^{4 / 3}$.

Let $t=\frac{(a+b+c)^{4}}{(a b c)^{4 / 3}}$. By the Cauchy-Schwarz inequality, $t \geqslant 81$, hence

$$
A \leqslant \frac{(a+b+c)^{4}-75(a b c)^{4 / 3}}{(a+b+c)^{4}-79(a b c)^{4 / 3}}=\frac{t-75}{t-79}=1+\frac{4}{t-79} \leqslant 3
$$

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