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

$$
A=\frac{a^{4}+b^{4}+c^{4}}{(a+b+c)^{4}-80(a b c)^{4 / 3}}
$$
Answer: 3.

Solution. Note that $(a+b+c)^{4}=a^{4}+b^{4}+c^{4}+B$, where

\[
\begin{aligned}
& B=4\left(a^{3} b+a b^{3}+b^{3} c+b c^{3}+c^{3} a+c a^{3}\right)+6\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+12\left(a b c^{2}+b c a^{2}+c a b^{2}\right) \geqslant \\
& \geqslant 24 \sqrt[6]{(a b c)^{8}}+18 \sqrt[3]{(a b c)^{4}}+36 \sqrt[3]{(a b c)^{4}}=78(a b c)^{4 / 3}
\end{aligned}
\]

(we used the Cauchy inequality). Then

\[
a^{4}+b^{4}+c^{4} \leqslant(a+b+c)^{4}-78(a b c)^{4 / 3}
\]

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

\[
A \leqslant \frac{(a+b+c)^{4}-78(a b c)^{4 / 3}}{(a+b+c)^{4}-80(a b c)^{4 / 3}}=\frac{t-78}{t-80}=1+\frac{2}{t-80} \leqslant 3
\]

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