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Daca \( a, b, c>0 \) astfel incat \( a^2+b^2+c^2=3 \), atunci avem \( \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{3}{2} \).

Cezar Lupu, lista scurta, 2008

\( \sum_{ciclic}{} {\frac {a^2}{b+c}} = \sum_{ciclic}{} {\frac {a^4}{a^2b+a^2c}} \) \( \frac {Cauchy}{\geq} \) \( \frac {{(a^2+b^2+c^2)}^2}{\sum_{ciclic}{} {a^2b+c^2b}} \) \( = \frac {3(a^2+b^2+c^2)}{\sum_{ciclic}{} {a(3-a^2)}} \)
\( \geq \frac {3}{2} \) \( \Leftrightarrow 2(a^2+b^2+c^2)+a^3+b^3+c^3 \geq 3(a+b+c) \) (1)
Dar avem: \( a^3+b^3+c^3 \geq \frac {(a^2+b^2+c^2)(a+b+c)}{3} = a+b+c \) (2)
\( 3=a^2+b^2+c^2 \geq \frac {{(a+b+c)}^2}{3} \) \( \Rightarrow 3 \geq a+b+c \Rightarrow a^2+b^2+c^2 \geq a+b+c \Rightarrow 2(a^2+b^2+c^2) \geq 2(a+b+c) \) (3)

Din (1) , (2) si (3) rezulta concluzia.

Tripletele \( (a^2,b^2,c^2) \) si \( (\frac{1}{b+c},\frac{1}{a+c},\frac{1}{a+b}) \) sunt la fel orientate si cu inegalitatea Cebasev avem
\( \sum\frac{a^2}{b+c} \)\( \geq \)\( (\sum \)\( a^2) \)\( (\sum \)\( \frac{1}{b+c}) \)\( \frac{1}{3} \)\( \geq \)\( \sum\frac{9}{2(a+b+c)} \)\( \geq \)\( \frac{3}{2} \) deoarece \( (a+b+c)^2 \)\( \leq3(\sum \)\( a^2) \).