Let a_1 ,a_2 ....a_n ∈R^+ , a_1 +a_2 +.....a_n =1 prove that: (a_1 /(2−a_1 ))+(a_2 /(2−a_2 )).......(a_n /(2−a


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Question Number 194037 by Subhi last updated on 26/Jun/23

$L e t a_{1}, a_{2} . . . . a_{n} \in R^{+}, a_{1} + a_{2} + . . . . . a_{n} = 1$ $p r o v e t h a t :$ $\frac{a_{1}}{2 - a_{1}} + \frac{a_{2}}{2 - a_{2}} . . . . . . . \frac{a_{n}}{2 - a_{n}} ⩾ \frac{n}{2 n - 1}$
	Answered by deleteduser1 last updated on 26/Jun/23

	$W L O G, L e t a_{1} ⩾ a_{2} ⩾ . . . ⩾ a_{n}$ $\Rightarrow \frac{1}{2 - a_{1}} ⩾ \frac{1}{2 - a_{2}} ⩾ . . . ⩾ \frac{1}{2 - a_{n}}$ $C h e b y s h e v \Rightarrow Σ \frac{a}{2 - a_{1}} ⩾ \frac{1}{n} (Σ a_{1}) (Σ \frac{1}{2 - a_{1}}) = \frac{1}{n} Σ (\frac{1}{2 - a_{1}})$ $⩾ \frac{1}{n} (\frac{n^{2}}{2 n - Σ a_{1}}) = \frac{n}{2 n - 1}$
	Commented by Subhi last updated on 26/Jun/23

	$p e r f e c t$
	Answered by Subhi last updated on 26/Jun/23

	$A n o t h e r s o l u t i o n$ $\frac{a_{1}^{2}}{2 a_{1} - a_{1}^{2}} + \frac{a_{2}^{2}}{2 a_{2} - a_{2}^{2}} . . . . . . \frac{a_{n}^{2}}{2 a_{n} - a_{n}^{2}} ⩾ \frac{{(a_{1} + a_{2} . . . a_{n})}^{2}}{2 (a_{1} + a_{2} . . . . . a_{n}) - (a_{1}^{2} + a_{2}^{2} . . . a_{n}^{2})}$ $⩾ \frac{1}{2 - \frac{{(a_{1} + a_{2} . . . . a_{n})}^{2}}{n}} = \frac{1}{2 - \frac{1}{n}} = \frac{n}{2 n - 1}$
	Answered by witcher3 last updated on 26/Jun/23

	$f (x) = \frac{1}{2 - x}, f^{'} (x) = \frac{1}{{(2 - x)}^{2}} > 0$ $f^{″} = \frac{2}{{(2 - x)}^{3}} ⩾ 0 f convex$ $\Rightarrow a_{i} f (a_{i}) ⩾ f (Σ a_{i}^{2}) . . . E$ $Quadratic Mean \Rightarrow \sqrt[2]{\frac{Σ a_{i}^{2}}{n}} ⩾ \frac{Σ a_{i}}{n}$ $Σ a_{i}^{2} ⩾ \frac{1}{n} . . . f increase function \Rightarrow f (Σ a_{i}^{2}) ⩾ f (\frac{1}{n})$ $E \Rightarrow \underset{i = 1}{\sum^{n}} \frac{a_{i}}{2 - a_{i}} ⩾ f (\frac{1}{n}) = \frac{n}{2 n - 1}$
	Commented by Subhi last updated on 26/Jun/23

	$n i c e! ({j e n s e n}^{'} s i n e q u a l i t y)$
	Commented by witcher3 last updated on 27/Jun/23

	$thank You sir Yes$
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