Conc. interj. "Gheorghe Lazar" Sibiu 2010 probl. 3

[Andi Brojbeanu]

Daca \( a, b, c\in [0,1] \) si \( ab+bc+ca=1 \), atunci demonstrati inegalitatea \( (a+b)(b+c)(c+a)<\frac{4}{3}(a+b+c) \).
Emil C. Popa, Sibiu

[Andi Brojbeanu]

\( (a+b)(b+c)(c+a)<\frac{2}{3}\cdot [(a+b)+(b+c)+(c+a)]\Leftrightarrow \frac{a+b+b+c+c+a}{(a+b)(b+c)(c+a)}>\frac{3}{2}\Leftrightarrow \sum \frac{1}{(a+b)(a+c)}>\frac{3}{2}\Leftrightarrow \sum\frac{1}{a^2+1}>\frac{3}{2} \).
Ultima inegalitate este adevarata intrucat \( a, b, c\in [0,1] \), deci cei trei termeni sunt \( \le \frac{1}{1^2+1}=\frac{1}{2} \). Egalitatea ar avea loc daca \( a=b=c=1 \), ceea ce contrazice ipoteza \( ab+bc+ca=1 \).