CF 1773F - Football
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 $abc=1$.
Rating: 800
Tags: constructive algorithms
Solve time: 2m 23s
Verified: no
Solution
Problem-Type Check
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.
Step-by-Step Verification
Step 1: Rewrite the expression using $abc=1$
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
$$\sum \frac1{a^3(b+c)} = \sum \frac{b^3c^3}{b+c}.$$
Step 2: Apply Engel-Cauchy
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)+(