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Fie \( a,b,c\in (0,\infty) \) astfel incat abc=1. Demonstrati ca:

\( (a+b)(b+c)(c+a)\ge 4(a+b+c-1). \)

Este binecunoscuta identitatea \( (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-1 \). Inegalitatea devine: \( \left(a+b+c\right)\left(ab+bc+ca\right)-1\geq4\left(a+b+c\right)-4\Longleftrightarrow ab+bc+ca+\frac{3}{a+b+c}\geq4 \). Insa
\( ab+bc+ca+\frac{3}{a+b+c}=\frac{ab+bc+ca}{3}+\frac{ab+bc+ca}{3}+\frac{ab+bc+ca}{3}+\frac{3}{a+b+c}. \)
Din inegalitatea mediilor avem \( \frac{ab+bc+ca}{3}+\frac{ab+bc+ca}{3}+\frac{ab+bc+ca}{3}+\frac{3}{a+b+c}\geq4\sqrt[4]{\frac{\left(ab+bc+ca\right)^{3}}{9\left(a+b+c\right)}}. \)
Mai ramane de demonstrat ca \( \left(ab+bc+ca\right)^{3}\geq9\left(a+b+c\right) \) , inegalitate care se obtine din inmultirea urmatoarelor inegalitati:
\( ab+bc+ca\geq3\sqrt[3]{a^{2}b^{2}c^{2}}=3 \) si \( \left(ab+bc+ca\right)^{2}\geq3abc\left(a+b+c\right)=3\left(a+b+c\right) \).