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Fie \( n \geq 4 \) natural si \( a_1, ... ,a_n \in R_+ \) astfel incat \( \sum_{i=1}^{n} {{a_i}^2} = 1 \). Aratati ca:

\( \sum_{k=1}^{n} {\frac {a_k}{{a_{k+1}}^2 + 1} \) \( \geq \) \( \frac {4}{5} \) \( {(\sum_{k=1}^{n} {a_k \sqrt {a_k}})}^2 \) unde \( a_{n+1}=a_1 \)


TST III - 2002 Problema 2

Folosim inegalitatea CBS


\( LHS=\sum {\frac{a_k^3}{a_k^2a_{k+1}^2+a_k^2}}\geq\frac{(\sum {a_k\sqrt{a_k}})^2}{\sum {a_k^2a_{k+1}^2+a_k^2}}\geq RHS \) deoarece

\( \sum_{k=1}^n {a_k^2a_{k+1}^2}\leq\frac{1}{4} \) in conditia ca \( \sum {a_k^2}=1 \)

Demonstrati inegalitatea .