Let a, b, c be positive numbers such that $abc=1$. Prove that $1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$
The usual methods do not seem to work, including a substitution $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ and trying to apply Muirhead's inequality.
Best Answer
We have $(ab+bc+ca)^2 \geq 3 (a+b+c)abc $, which is equivalent to $ (ab-bc)^2 + (bc-ca)^2 + (ca-ab)^2 \geq 0$. Hence,
$$ 1 + \frac{3}{a+b+c} \geq 2 \sqrt{ \frac{3}{a+b+c} } \geq \frac{6}{ab+bc+ca}. $$