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Daca a,b,c >0 si \( a^2+b^2+c^2=1 \) atunci

a) \( \frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\ge \frac{3\sqrt{3}}{3} \)

b) \( \frac{a}{1-a^4}+\frac{b}{1-b^4}+\frac{c}{1-c^4}\ge \frac{5\sqrt[4]{5}}{4} \)

a) Relatia este echivalenta cu:

\( \sum\limits_{cyc} {\frac{{a^2 + b^2 }}{c}} \ge \sqrt 3 \)

\( \frac{1}{c} \le \frac{1}{b} \le \frac{1}{a}{\rm \ si\ a}^{\rm 2} + b^2 \le a^2 + c^2 \le b^2 + c^2 \)

Aplicam Cebasev:

\( \frac{2}{3}\left( {\frac{1}{c} + \frac{1}{b} + \frac{1}{a}} \right)\left( {a^2 + b^2 + c^2 } \right) \le \sum\limits_{cyc} {\frac{{{\rm a}^{\rm 2} + b^2 }}{c}}\Rightarrow \sum\limits_{cyc} {\frac{{{\rm a}^{\rm 2} + b^2 }}{c}} \ge \frac{2}{3}\left( {\sum {\frac{1}{a}} } \right) \)

Aplicam acum CBS:
\( \frac{2}{3}\left( {\frac{1}{c} + \frac{1}{b} + \frac{1}{a}} \right) \ge \frac{2}{3}\cdot 9\cdot \frac{1}{{a + b + c}} \ge \frac{6}{{\sqrt {3\left( {a^2 + b^2 + c^2 } \right)} }} = 2\sqrt 3 \ge \sqrt 3 \)