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Problema 5, lista scurta 2009

Posted: Sun Apr 19, 2009 8:57 pm
by alex2008
a) Daca \( a,b\in (0,\infty ) \) si \( ab=1 \), aratati ca :

\( \frac{a^2}{a+1}+\frac{b^2}{b+1}\ge 1 \)

b) Fie \( a_1,a_2,...,a_n \in (0,\infty )\ ,\ n\in \mathbb{N}\ ,\ n\ge 2 \) , astfel incat \( a_1a_2\cdot ...\cdot a_n=1 \). Daca \( a_1+a_2+...+a_n=s \), aratati ca :

\( \frac{a_1^2}{s+1-a_1}+\frac{a_2^2}{s+1-a_2}+...+\frac{a_n^2}{s+1-a_n}\ge 1 \)

Traian Tamaian, Carei

Posted: Mon Apr 20, 2009 2:48 pm
by Marius Mainea
Si la a) si la b) se aplica C-B-S.

b) \( LHS\ge \frac{(a_1+a_2+...+a_n)^2}{(n-1)(a_1+a_2+...+a_n)+n}\ge 1 \) deoarece \( s\ge \sqrt[n]{a_1a_2...a_n}=1 \).