Inegalitate

[Claudiu Mindrila]

Aratati ca pentru orice \( x,y,z\in\left(0,\infty\right) \) are loc inegalitatea: \( \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\ge\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{z+x}{z+x+2y} \).

Marin Chirciu, R.M.T. 1/2009

[Marius Mainea]

Notand y+z=a , z+x=b , x+y=c , inegalitatea este echivalenta cu

\( \sum {\frac{b+c-a}{2a}}\ge \sum {\frac{c}{a+b}} \)

Insa folosind AM-HM \( RHS\le \frac{1}{4}\sum {a(\frac{1}{b}+\frac{1}{c})}\le LHS \) deoarece

\( \sum {\frac{a+b}{c}}\ge \6 \)

[baleanuAR]

Cateva idei si aici http://www.mathlinks.ro/Forum/viewtopic ... 30#1425030