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Question Number 193875 by Subhi last updated on 22/Jun/23

a, b, c, d, e, f, a r e + r e a l n u m b e r s

p r o v e :

\frac{a}{b + c} + \frac{b}{c + d} + \frac{c}{d + e} + \frac{d}{e + f} + \frac{e}{f + a} + \frac{f}{a + b} ⩾ 3

Answered by AST last updated on 23/Jun/23

= \sum_{c y c} \frac{a^{2}}{a (b + c)} ⩾ \frac{{(Σ a)}^{2}}{\sum_{c y c} a (b + c)} \dots . (i)

2 \sum_{c y c} a (b + c) = {(Σ a)}^{2} - {(a + d)}^{2} - {(b + e)}^{2} - {(c + f)}^{2}

F r o m 3 (a^{2} + b^{2} + c^{2}) ⩾ {(a + b + c)}^{2}

W e g e t 2 \sum_{c y c} a (b + c) ⩽ \frac{2}{3} {(Σ a)}^{2} \dots (i i)

(i) a n d (i i) \Rightarrow Σ \frac{a^{2}}{a (b + c)} ⩾ \frac{2 {(Σ a)}^{2}}{2 Σ a (b + c)} ⩾ 3

Commented by Subhi last updated on 23/Jun/23

p e r f e c t

Answered by Subhi last updated on 23/Jun/23

\sum_{c y c} a (b + c) \sum_{c y c} \frac{(a)}{(b + c)} ⩾ {(Σ a)}^{2}

Σ \frac{(a)}{(b + c)} ⩾ \frac{{(Σ a)}^{2}}{Σ a (b + c)} = \frac{{(a + b + c + d + e + f)}^{2}}{a b + a c + b c + b d + c d + c e + d e + d f + e f + e a + a f + f b}

2 Σ a (b + c) = {(\sum_{c y c} a)}^{2} - Σ a^{2} - 2 (a d + b e + c f) = {(\sum_{c y c} a)}^{2} - ({(a + d)}^{2} + {(b + e)}^{2} + {(c + f)}^{2})

{(a + d)}^{2} + {(b + e)}^{2} + {(c + f)}^{2} ⩾ \frac{{(a + b + c + d + e + f)}^{2}}{3}

2 Σ a (b + c) ⩽ \frac{2}{3} {(Σ a)}^{2} ⇛ Σ a (b + c) ⩽ \frac{1}{3} {(Σ a)}^{2}

∴ \sum_{c y c} \frac{a}{b + c} ⩾ \frac{3 {(Σ a)}^{2}}{{(Σ a)}^{2}} = 3

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