Competition problem — Algebra — Maths Olympiad Prep

Library / /116 of 520

Algebra Difficulty 6.1 National olympiad Prove it

Example 8.2.3 Let $a, b, c>0$ , and satisfy $abc=1$ , prove:
$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a} \leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c} \quad$ (Bulgarian MO 1998)

Solution

Proof: Let $S=\sum_{cyc} a, P=\sum_{cyc} ab, Q=abc$ , through not too complicated calculations, we have
$LHS=\sum_{cyc} \frac{1}{S+1-a}=\frac{S^{2}+4S+3+P}{S^{2}+2S+PS+P}, \quad RHS=\sum_{cyc} \frac{1}{2+a}=\frac{12+4S+P}{9+4S+2P}$

Therefore, we only need to prove
$\frac{S^{2}+4S+3+P}{S^{2}+2S+PS+P} \leq \frac{12+4S+P}{9+4S+2P}, \text{ i.e., } (3P-5)S^{2}+(S-1)P^{2}+6PS \geq 24S+3P+27$

Since $abc=1$ , we have $S, P \geq 3$ , so
$LHS \geq 4S^{2}+2P^{2}+6PS \geq 12S+6(P-1)S+6S+2P^{2} \geq 24S+3P+(P^{2}+6S) \geq RHS$

The equality holds when $S=P=3$ or $a=b=c=1$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.