CF 1769D3 - Игра в Девятку III

Rating: 2300
Tags: *special, brute force
Solve time: 2m 14s
Verified: no

Solution

This is a Type B (prove inequality) problem, not Type C.

The task is to prove that

$$\frac1{a^3(b+c)}+\frac1{b^3(c+a)}+\frac1{c^3(a+b)}\ge \frac32$$

for all positive $a,b,c$ satisfying $abc=1$.

A valid solution must establish the inequality for all admissible triples and correctly identify any equality case if it arises from the argument.

The solution claims

$$\frac1{a^3(b+c)}=\frac{b^3c^3}{b+c},$$

and similarly for the cyclic terms.

Assessment: VALID.

Since $abc=1$, we have $a=\frac1{bc}$, hence $a^{-3}=b^3c^3$.

Thus

$$\sum \frac1{a^3(b+c)} = \sum \frac{b^3c^3}{b+c}.$$

The solution applies

\sum \frac{(b^{3/2}c^{3/2})^2}{b+c} \ge \frac{\left(\sum b^{3/2}c^{3/2}\right)^2} {(a+b)+(b+c)+(