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

$$
A=\frac{a^{3}+b^{3}+c^{3}}{(a+b+c)^{3}-26 a b c}
$$
Answer: 3.

Solution. 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 inequality). Then

$$
a^{3}+b^{3}+c^{3} \leqslant(a+b+c)^{3}-24 a b c
$$

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

$$
A \leqslant \frac{(a+b+c)^{3}-24 a b c}{(a+b+c)^{3}-26 a b c}=\frac{t-24}{t-26}=1+\frac{2}{t-26} \leqslant 3
$$

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