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-n-n-2n-1-

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

L e t a_{1}, a_{2} \dots . a_{n} \in R^{+}, a_{1} + a_{2} + \dots . . a_{n} = 1

p r o v e t h a t :

\frac{a_{1}}{2 - a_{1}} + \frac{a_{2}}{2 - a_{2}} \dots \dots . \frac{a_{n}}{2 - a_{n}} ⩾ \frac{n}{2 n - 1}

Answered by AST last updated on 26/Jun/23

W L O G, L e t a_{1} ⩾ a_{2} ⩾ \dots ⩾ a_{n}

\Rightarrow \frac{1}{2 - a_{1}} ⩾ \frac{1}{2 - a_{2}} ⩾ \dots ⩾ \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}} \dots \dots \frac{a_{n}^{2}}{2 a_{n} - a_{n}^{2}} ⩾ \frac{{(a_{1} + a_{2} \dots a_{n})}^{2}}{2 (a_{1} + a_{2} \dots . . a_{n}) - (a_{1}^{2} + a_{2}^{2} \dots a_{n}^{2})}

⩾ \frac{1}{2 - \frac{{(a_{1} + a_{2} \dots . 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}) \dots E

Quadratic Mean \Rightarrow \sqrt[2]{\frac{Σ a_{i}^{2}}{n}} ⩾ \frac{Σ a_{i}}{n}

Σ a_{i}^{2} ⩾ \frac{1}{n} \dots 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