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Question Number 193707 by Mingma last updated on 18/Jun/23

	Answered by Subhi last updated on 18/Jun/23

	$\frac{a^{2}}{b^{2} + c^{2}} + \frac{b^{2}}{a^{2} + c^{2}} + \frac{c^{2}}{a^{2} + b^{2}} + 3 = \sum_{c y c} \frac{a^{2} + b^{2} + c^{2}}{b^{2} + c^{2}}$ $2 (a^{2} + b^{2} + c^{2}) (\frac{1}{a^{2} + b^{2}} + \frac{1}{a^{2} + c^{2}} + \frac{1}{b^{2} + c^{2}}) ⩾ {(1 + 1 + 1)}^{2} = 9 (c h a u c h y_s c h w a r t z)$ $∴ \sum_{c y c} \frac{a^{2}}{b^{2} + c^{2}} ⩾ \frac{9}{2} - 3 = \frac{3}{2}$ $∴ {c o s}^{2} (α) + {c o s}^{2} (β) + {c o s}^{2} (γ) ⩾ \frac{3}{4} (p r o o f)$ ${c o s}^{2} (α) = \frac{c o s (2 α) + 1}{2}$ $\frac{c o s (2 α) + c o s (2 β) + c o s (2 γ) + 3}{2} ⩾ \frac{3}{4}$ $c o s (2 α) + c o s (2 β) = 2 c o s (α + β) c o s (α - β)$ $2 c o s (α + β) c o s (α - β) + 2 {c o s}^{2} (γ) + 2 ⩾ \frac{3}{2} (r e q u i r e d)$ $α + β = 180 - γ ⇛ c o s (α + β) = - c o s (γ)$ $2 {c o s}^{2} (γ) - 2 c o s (α - β) c o s (γ) + 2$ $2 {(c o s (γ) - \frac{c o s (α - β)}{2})}^{2} - \frac{{c o s}^{2} (α - β)}{2} + 2$ ${c o s}^{2} (α - β) = 1 - {s i n}^{2} (α - β)$ $2 {(c o s (γ) - \frac{c o s (α - β)}{2})}^{2} + \frac{{s i n}^{2} (α - β)}{2} - \frac{1}{2} + 2$ $2 {(c o s (γ) - \frac{c o s (α - β)}{2})}^{2} + \frac{{s i n}^{2} (α - β)}{2} + \frac{3}{2} ⩾ \frac{3}{2}$ $n o t e t h a t {s i n}^{2}, {()}^{2} ⩾ 0$ $∴ 2 \sum_{c y c} {c o s}^{2} (α) ⩾ \sum_{c y c} \frac{a^{2}}{b^{2} + c^{2}}$
	Commented by Mingma last updated on 18/Jun/23
	Ver nice solution, sir!
	Commented by York12 last updated on 04/Mar/24

	$∴ \sum_{c y c} \frac{a^{2}}{b^{2} + c^{2}} ⩾ \frac{9}{2} - 3 = \frac{3}{2} how is that useful$
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