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Question Number 97968 by HamraboyevFarruxjon last updated on 11/Jun/20


Commented by 06122004 last updated on 10/Jun/20
Osonmasmi ��
	Answered by 1549442205 last updated on 11/Jun/20

	$Applying {Bunhiacopxky}^{'} s inequality :$ ${(a_{1} x_{1} + a_{2} x_{2} + . . . + a_{n} x_{n})}^{2} ⩽ (a_{1}^{2} + a_{2}^{2} + . . . + a_{n}^{2}) (x_{1}^{2} + x_{2}^{2} + . . . + x_{n}^{2}) we have$ ${(a_{1} + a_{2} + . . . + a_{n})}^{2} = {(\sqrt{\frac{a_{1}}{a_{2} + a_{3}}} . \sqrt{a_{1} (a_{2} + a_{3})} + \sqrt{\frac{a_{2}}{a_{3} + a_{4}}} . \sqrt{a_{2} (a_{3} + a_{4})} + . . . + \sqrt{\frac{a_{n}}{a_{1} + a_{2}}} . \sqrt{a_{n} (a_{1} + a_{2})})}^{2}$ $⩽ (\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + . . . + \frac{a_{n}}{a_{1} + a_{2}}) (a_{1} a_{2} + a_{1} a_{3} + . . . + a_{n} a_{1} + a_{n} a_{2})$ $we get \frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + . . . + \frac{a_{n}}{a_{1} + a_{2}} ⩾ \frac{{(a_{1} + a_{1} + . . . + a_{n})}^{2}}{a_{1} a_{2} + a_{1} a_{3} + . . . + a_{n} a_{1} + a_{n} a_{2}} .$ $Now we need to prove that$ $\frac{{(a_{1} + a_{1} + . . . + a_{n})}^{2}}{a_{1} a_{2} + a_{1} a_{3} + . . . + a_{n} a_{1} + a_{n} a_{2}} ⩾ \frac{n}{2} \Leftrightarrow 2 {(a_{1} + a_{1} + . . . + a_{n})}^{2} ⩾ n (a_{1} a_{2} + a_{1} a_{3} + . . . + a_{n} a_{1} + a_{n} a_{2})$ $\Leftrightarrow 2 (a_{1}^{2} + a_{2}^{2} + . . . + a_{n}^{2}) - 2 \underset{i \neq j}{Σ} a_{i} a_{j} = {(a_{1} - a_{2})}^{2} + {(a_{2} - a_{3})}^{2} + . . . + {(a_{n} - a_{1})}^{2} ⩾ 0$ $The equlity sign occurs if and only if$ $a_{1} = a_{2} = . . . = a_{n} (q . e . d)$
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