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Inegalitate usoara

Posted: Sat Jul 11, 2009 10:02 pm
by Claudiu Mindrila
\( \left\{ \begin{array}{c}
a_{1},\ a_{2},\dots,\ a_{n}\in\left(0,\ +\infty\right)\\
S=\sum_{k=1}^{n}a_{k}\\
\ a_{n+1}=a_{1}\end{array}\right|\Longrightarrow\sum_{k=1}^{n}\sqrt{\frac{a_{k}+a_{k+1}}{2S-a_{k}-a_{k+1}}}\ge2 \)


Dumitru Acu

Posted: Sun Jul 12, 2009 7:19 pm
by Mateescu Constantin
Din \( MH-MG \) avem:

\( \frac{2}{1+\frac{2S-a_k-a_{k+1}}{a_k+a_{k+1}}}=\frac{a_k+a_{k+1}}{S}\le \sqrt{\frac{a_k+a_{k+1}}{2S-a_k-a_{k+1}}\ \cdot\ 1} \)

\( \Longrightarrow\ LHS\ge \sum_{k=1}^n \frac{a_k+a_{k+1}}{S}=\frac{2S}{S}=2 \).