inegalitate rationala

[Marius Mainea]

Daca x,y,z sunt pozitive atunci

\( \frac{x}{y^2+yz+z^2}+\frac{y}{z^2+zx+x^2}+\frac{z}{x^2+xy+y^2}\ge\frac{3}{x+y+z} \)

G.M.2009

[mihai++]

\( \sum\frac{x}{y^2+yz+z^2}\geq \frac{(x+y+z)^2}{\sum x(y^2+yz+z^2)}=\frac{(x+y+z)^2}{(x+y+z)(xy+yz+zx)}\geq\frac{3}{x+y+z} \).