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Inegalitate
Posted: Thu Feb 26, 2009 9:16 am
by 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
Posted: Thu Feb 26, 2009 10:31 am
by 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
\( \sum {\frac{a+b}{c}}\ge \6 \)
Posted: Tue Mar 03, 2009 4:39 pm
by baleanuAR
