Question

Problema 4. Multumesc!

Answer 1

$\displaystyle Doar~notezi~b_i=\ln a_i.~i=\overline{1,n}. \\ \\ Evident~b_1\ \textgreater \ 0,~si~notand~S=b_1+b_2+...+b_n,~inegalitatea~devine~ \\ \\ \frac{b_1}{S-b_1}+ \frac{b_2}{S-b_2}+...+ \frac{b_n}{S-b_n} \ge \frac{n-1}{n}.~(*) \\ \\ Aduni~n=1+1+...+1~in~ambii~membri: \\ \\ (*) \Leftrightarrow \sum\limits_{k=1}^n \frac{S}{S-b_k} \ge n+ \frac{n}{n-1} \Leftrightarrow S\sum\limits_{k=1}^n \frac{1}{S-b_k} \ge \frac{n^2}{n-1} . \\ \\ Din~CBS,~membrul~stang~este \ge S \cdot \frac{n^2}{nS-(b_1+b_2+...+b_n)}=$
$\displaystyle = \frac{n^2S}{(n-1)S}= \frac{n^2}{n-1}.~Q.E.D.$