Prove that: In any acute △ABC, cot^2 A+cot^2 B+cot^2 C≥1. Equality is possible if and only if A=B=C=(π/3).


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Question Number 193237 by CrispyXYZ last updated on 08/Jun/23

$Prove that :$ $In any acute △ A B C, \cot^{2} A + \cot^{2} B + \cot^{2} C ⩾ 1 .$ $Equality is possible if and only if A = B = C = \frac{π}{3} .$
	Answered by MM42 last updated on 09/Jun/23

	$0 < α, β, γ < 90 \Rightarrow c o t α, c o t β, c o t γ > 0$ ${c o t}^{2} α + {c o t}^{2} β + {c o t}^{2} γ =$ $\frac{{c o t}^{2} α + {c o t}^{2} β}{2} + \frac{{c o t}^{2} α + {c o t}^{2} γ}{2} + \frac{{c o t}^{2} β + {c o t}^{2} γ}{2} ⩾$ $\sqrt{{c o t}^{2} α \times {c o t}^{2} β} + \sqrt{{c o t}^{2} α \times {c o t}^{2} γ} + \sqrt{{c o t}^{2} γ \times {c o t}^{2} β} =$ $c o t α (c o t β + c o t γ) + c o t β c o t γ = (0 < β, γ < \frac{π}{2} \to c o t β \times c o t γ \neq 1)$ $c o t α (\frac{c o t β + c o t γ}{c o t β c o t γ - 1}) (c o t β c o t γ - 1) + c o t β \times c o t γ =$ $c o t α \times \frac{1}{c o t (β + γ)} (c o t β c o t γ - 1) + c o t β c o t γ = ()$ $β + γ = π - α$ $\Rightarrow () = - c o t β c o t β + 1 + c o t β c o t γ = 1 ✓$ $w e s h o w e d : c o t α \times c o t β + c o t α \times c o t γ + c o t β \times c o t γ = 1$ $\Rightarrow i f c o \overset{2}{t} α + {c o t}^{2} β + {c o t}^{2} γ = 1$ $c o \overset{2}{t} α + {c o t}^{2} β + {c o t}^{2} γ = \frac{1}{2} [{(c o t α - c o t β)}^{2} + {(c o t α - c o t γ)}^{2} + {(c o t β - c o t γ)}^{2}] + c o t α \times c o t β + c o t α \times c o t γ + c o t β \times c o t γ$ $\Rightarrow \frac{1}{2} [{(c o t α - c o t β)}^{2} + {(c o t α - c o t γ)}^{2} + {(c o t β - c o t γ)}^{2}] = 0$ $\Rightarrow c o t α = c o t β = c o t γ \Rightarrow α = β = γ = \frac{π}{3} ✓$ $i f α = β = γ = \frac{π}{4} \Rightarrow {c o t}^{2} α + {c o t}^{2} β + {c o t}^{2} γ = 1 ✓$
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