Question

[tex]Fie~a_1,a_2,...,a_n\in[0,1]~cu~\sum a_n=1,n\geq 2.Sa~se~dem~ca:\\ \sum\limits_{k=1}^n \sqrt{a_k(1-a_k)} \leq \sqrt{n-1}.\\ Am~obs~inca~din~conditiile~a_k \in[0,1] , k= \overline {1,n}~ca:\\ 0\leq a_k\leq 1/\cdot(1-a_k) \Rightarrow 0\leq a_k(1-a_k)\leq 1-a_k~si~deci~\\ 0\leq \sqrt{a_k(1-a_k)}\leq \sqrt{1-a_k}.\\ Ca~urmare~\sum\limits_{k=1}^n \sqrt{a_k(1-a_k)}\leq \sum\limits_{k=1}^n \sqrt{1-a_k}\leq \sqrt{n-1} [/tex]

Am incercat cu AM-GM dar nu merge sa dem ultima parte...asta a fost ideea mea.M-am mai gandit si la CBS dar nu am incercat..ma puteti ajuta?

Answer 1

[tex]\text{Merge cu CBS:}\\ \left(\displaystyle\sum_{k=1}^n(\sqrt{a_k})^2\right)\left(\displaystyle\sum_{k=1}^n(\sqrt{1-a_k})^2\right)\\ \text{Mai departe avem ca:}\\ \left(\displaystyle\sum_{k=1}^n a_k\right)\left(\displaystyle\sum_{k=1}^n (1-a_k)\right)\geq \left( \displaystyle\sum_{k=1}^n\sqrt{a_k(1-a_k)}\right)^2\\ [/tex]
$\underbrace{\left(\displaystyle\sum_{k=1}^n a_k\right)}\left(n-\underbrace{\left(\displaystyle\sum_{k=1}^n a_k\right)}\right)\geq \left( \displaystyle\sum_{k=1}^n\sqrt{a_k(1-a_k)}\right)^2\\ ~~~~~~=1~~~~~~~~~~~~~~=1\\ 1(n-1)\geq \left( \displaystyle\sum_{k=1}^n\sqrt{a_k(1-a_k)}\right)^2\\ \text{Deci:}\\ \displaystyle\sum_{k=1}^n\sqrt{a_k(1-a_k)}\leq \sqrt{n-1},Q.E.D.$