Mathematical Reflections 2013, Issue 1 - Problem J253

Problem:
Prove that if $a,b,c>0$ satisfy $abc=1$, then $$\dfrac{1}{ab+a+2}+\dfrac{1}{bc+b+2}+\dfrac{1}{ca+c+2} \leq \dfrac{3}{4}.$$

Proposed by Marcel Chirita.

Solution:
By the AM-HM Inequality, we have $$\dfrac{1}{ab+1+a+1} \leq \dfrac{1}{4}\left(\dfrac{1}{ab+1}+\dfrac{1}{a+1}\right)=\dfrac{1}{4}\left(\dfrac{c}{c+1}+\dfrac{1}{a+1}\right)$$ $$\dfrac{1}{bc+1+b+1} \leq \dfrac{1}{4}\left(\dfrac{1}{bc+1}+\dfrac{1}{b+1}\right)=\dfrac{1}{4}\left(\dfrac{a}{a+1}+\dfrac{1}{b+1}\right)$$ $$\dfrac{1}{ca+1+c+1} \leq \dfrac{1}{4}\left(\dfrac{1}{ca+1}+\dfrac{1}{c+1}\right)=\dfrac{1}{4}\left(\dfrac{b}{b+1}+\dfrac{1}{c+1}\right).$$
Summing up the three inequalities, we obtain $$\dfrac{1}{ab+a+2}+\dfrac{1}{bc+b+2}+\dfrac{1}{ca+c+2} \leq \dfrac{1}{4} \left( \dfrac{a+1}{a+1}+\dfrac{b+1}{b+1}+\dfrac{c+1}{c+1}\right)=\dfrac{3}{4}.$$