Inegalitate 3

[Marius Mainea]

Fie a,b,c pozitive. Demonstrati inegalitatea:

\( \frac{a^3+abc}{b+c}+\frac{b^3+abc}{c+a}+\frac{c^3+abc}{a+b}\ge a^2+b^2+c^2. \)

Cezar Lupu

[mihai++]

\( \sum \frac{a^3+abc}{b+c}\geq\sum a^2\Leftrightarrow\sum\frac{a(a-b)(a-c)}{b+c}\geq 0 \).
Dar \( \sum\frac{a(a-b)(a-c)}{b+c}\geq \frac{\sum a(a-b)(a-c)}{a+b+c}\geq 0 \), din Schur. Egalitatea se atinge cand \( a=b=c \).