Question Number 193237 by CrispyXYZ last updated on 08/Jun/23
$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{In}\:\mathrm{any}\:\mathrm{acute}\:\bigtriangleup{ABC},\:\mathrm{cot}^{\mathrm{2}} {A}+\mathrm{cot}^{\mathrm{2}} {B}+\mathrm{cot}^{\mathrm{2}} {C}\geqslant\mathrm{1}. \\ $$$$\mathrm{Equality}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:{A}={B}={C}=\frac{\pi}{\mathrm{3}}. \\ $$
Answered by MM42 last updated on 09/Jun/23
![0<α,β,γ<90⇒cotα ,cotβ,cotγ >0 cot^2 α+cot^2 β+cot^2 γ= ((cot^2 α+cot^2 β)/2) +((cot^2 α+cot^2 γ)/2)+((cot^2 β+cot^2 γ)/2)≥ (√(cot^2 α×cot^2 β)) +(√(cot^2 α×cot^2 γ)) +(√(cot^2 γ×cot^2 β))= cotα(cotβ+cotγ)+cotβcotγ= (0<β,γ<(π/(2 )) → cotβ×cotγ≠1 ) cotα(((cotβ+cotγ)/(cotβcotγ−1)))(cotβcotγ−1)+cotβ×cotγ= cotα×(1/(cot(β+γ)))(cotβcotγ−1)+cotβcotγ= (∗) β+γ=π−α ⇒(∗)=−cotβcotβ+1+cotβcotγ=1 ✓ we showed : cotα×cotβ+cotα×cotγ+cotβ×cotγ=1 ⇒if cot^2 α+cot^2 β+cot^2 γ=1 cot^2 α+cot^2 β+cot^2 γ=(1/2)[(cotα−cotβ)^2 +(cotα−cotγ)^2 +(cotβ−cotγ)^2 ]+ cotα×cotβ+cotα×cotγ+cotβ×cotγ ⇒(1/2)[(cotα−cotβ)^2 +(cotα−cotγ)^2 +(cotβ−cotγ)^2 ]=0 ⇒cotα=cotβ=cotγ⇒α=β=γ=(π/3) ✓ if α=β=γ=(π/4)⇒cot^2 α+cot^2 β+cot^2 γ=1 ✓](Q193274.png)
$$\mathrm{0}<\alpha,\beta,\gamma<\mathrm{90}\Rightarrow{cot}\alpha\:,{cot}\beta,{cot}\gamma\:>\mathrm{0} \\ $$$${cot}^{\mathrm{2}} \alpha+{cot}^{\mathrm{2}} \beta+{cot}^{\mathrm{2}} \gamma= \\ $$$$\frac{{cot}^{\mathrm{2}} \alpha+{cot}^{\mathrm{2}} \beta}{\mathrm{2}}\:+\frac{{cot}^{\mathrm{2}} \alpha+{cot}^{\mathrm{2}} \gamma}{\mathrm{2}}+\frac{{cot}^{\mathrm{2}} \beta+{cot}^{\mathrm{2}} \gamma}{\mathrm{2}}\geqslant \\ $$$$\sqrt{{cot}^{\mathrm{2}} \alpha×{cot}^{\mathrm{2}} \beta}\:+\sqrt{{cot}^{\mathrm{2}} \alpha×{cot}^{\mathrm{2}} \gamma}\:+\sqrt{{cot}^{\mathrm{2}} \gamma×{cot}^{\mathrm{2}} \beta}= \\ $$$${cot}\alpha\left({cot}\beta+{cot}\gamma\right)+{cot}\beta{cot}\gamma=\:\:\:\:\:\:\left(\mathrm{0}<\beta,\gamma<\frac{\pi}{\mathrm{2}\:}\:\rightarrow\:{cot}\beta×{cot}\gamma\neq\mathrm{1}\:\right) \\ $$$${cot}\alpha\left(\frac{{cot}\beta+{cot}\gamma}{{cot}\beta{cot}\gamma−\mathrm{1}}\right)\left({cot}\beta{cot}\gamma−\mathrm{1}\right)+{cot}\beta×{cot}\gamma= \\ $$$${cot}\alpha×\frac{\mathrm{1}}{{cot}\left(\beta+\gamma\right)}\left({cot}\beta{cot}\gamma−\mathrm{1}\right)+{cot}\beta{cot}\gamma=\:\:\left(\ast\right) \\ $$$$\beta+\gamma=\pi−\alpha \\ $$$$\Rightarrow\left(\ast\right)=−{cot}\beta{cot}\beta+\mathrm{1}+{cot}\beta{cot}\gamma=\mathrm{1}\:\checkmark\: \\ $$$${we}\:{showed}\::\:{cot}\alpha×{cot}\beta+{cot}\alpha×{cot}\gamma+{cot}\beta×{cot}\gamma=\mathrm{1} \\ $$$$\Rightarrow{if}\:\:\:\:{co}\overset{\mathrm{2}} {{t}}\alpha+{cot}^{\mathrm{2}} \beta+{cot}^{\mathrm{2}} \gamma=\mathrm{1} \\ $$$${co}\overset{\mathrm{2}} {{t}}\alpha+{cot}^{\mathrm{2}} \beta+{cot}^{\mathrm{2}} \gamma=\frac{\mathrm{1}}{\mathrm{2}}\left[\left({cot}\alpha−{cot}\beta\right)^{\mathrm{2}} +\left({cot}\alpha−{cot}\gamma\right)^{\mathrm{2}} +\left({cot}\beta−{cot}\gamma\right)^{\mathrm{2}} \right]+\:{cot}\alpha×{cot}\beta+{cot}\alpha×{cot}\gamma+{cot}\beta×{cot}\gamma \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left[\left({cot}\alpha−{cot}\beta\right)^{\mathrm{2}} +\left({cot}\alpha−{cot}\gamma\right)^{\mathrm{2}} +\left({cot}\beta−{cot}\gamma\right)^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\Rightarrow{cot}\alpha={cot}\beta={cot}\gamma\Rightarrow\alpha=\beta=\gamma=\frac{\pi}{\mathrm{3}}\:\checkmark \\ $$$${if}\:\:\alpha=\beta=\gamma=\frac{\pi}{\mathrm{4}}\Rightarrow{cot}^{\mathrm{2}} \alpha+{cot}^{\mathrm{2}} \beta+{cot}^{\mathrm{2}} \gamma=\mathrm{1}\:\checkmark \\ $$$$ \\ $$$$ \\ $$