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Question-193707

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Question Number 193707 by Mingma last updated on 18/Jun/23
Answered by Subhi last updated on 18/Jun/23
(a^2 /(b^2 +c^2 ))+(b^2 /(a^2 +c^2 ))+(c^2 /(a^2 +b^2 ))+3=Σ_(cyc) ((a^2 +b^2 +c^2 )/(b^2 +c^2 )) 2(a^2 +b^2 +c^2 )((1/(a^2 +b^2 ))+(1/(a^2 +c^2 ))+(1/(b^2 +c^2 )))≥(1+1+1)^2 =9 (chauchy_schwartz) ∴ Σ_(cyc) (a^2 /(b^2 +c^2 ))≥(9/2)−3=(3/2) ∴ cos^2 (α)+cos^2 (β)+cos^2 (γ)≥(3/4) (proof) cos^2 (α)=((cos(2α)+1)/2) ((cos(2α)+cos(2β)+cos(2γ)+3)/2)≥(3/4) cos(2α)+cos(2β)=2cos(α+β)cos(α−β) 2cos(α+β)cos(α−β)+2cos^2 (γ)+2≥(3/2) (required) α+β=180−γ ⇛ cos(α+β)=−cos(γ) 2cos^2 (γ)−2cos(α−β)cos(γ)+2 2(cos(γ)−((cos(α−β))/2))^2 −((cos^2 (α−β))/2)+2 cos^2 (α−β)=1−sin^2 (α−β) 2(cos(γ)−((cos(α−β))/2))^2 +((sin^2 (α−β))/2)−(1/2)+2 2(cos(γ)−((cos(α−β))/2))^2 +((sin^2 (α−β))/2)+(3/2)≥(3/2) note that sin^2 , ( )^2 ≥0 ∴ 2Σ_(cyc) cos^2 (α)≥Σ_(cyc) (a^2 /(b^2 +c^2 ))
$$\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\mathrm{3}=\underset{{cyc}} {\sum}\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\right)\geqslant\left(\mathrm{1}+\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{9}\:\left({chauchy\_schwartz}\right) \\ $$$$\therefore\:\underset{{cyc}} {\sum}\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\geqslant\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{3}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\therefore\:{cos}^{\mathrm{2}} \left(\alpha\right)+{cos}^{\mathrm{2}} \left(\beta\right)+{cos}^{\mathrm{2}} \left(\gamma\right)\geqslant\frac{\mathrm{3}}{\mathrm{4}}\:\left({proof}\right) \\ $$$${cos}^{\mathrm{2}} \left(\alpha\right)=\frac{{cos}\left(\mathrm{2}\alpha\right)+\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{cos}\left(\mathrm{2}\alpha\right)+{cos}\left(\mathrm{2}\beta\right)+{cos}\left(\mathrm{2}\gamma\right)+\mathrm{3}}{\mathrm{2}}\geqslant\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${cos}\left(\mathrm{2}\alpha\right)+{cos}\left(\mathrm{2}\beta\right)=\mathrm{2}{cos}\left(\alpha+\beta\right){cos}\left(\alpha−\beta\right) \\ $$$$\mathrm{2}{cos}\left(\alpha+\beta\right){cos}\left(\alpha−\beta\right)+\mathrm{2}{cos}^{\mathrm{2}} \left(\gamma\right)+\mathrm{2}\geqslant\frac{\mathrm{3}}{\mathrm{2}}\:\left({required}\right) \\ $$$$\alpha+\beta=\mathrm{180}−\gamma\:\Rrightarrow\:{cos}\left(\alpha+\beta\right)=−{cos}\left(\gamma\right) \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} \left(\gamma\right)−\mathrm{2}{cos}\left(\alpha−\beta\right){cos}\left(\gamma\right)+\mathrm{2} \\ $$$$\mathrm{2}\left({cos}\left(\gamma\right)−\frac{{cos}\left(\alpha−\beta\right)}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{{cos}^{\mathrm{2}} \left(\alpha−\beta\right)}{\mathrm{2}}+\mathrm{2} \\ $$$${cos}^{\mathrm{2}} \left(\alpha−\beta\right)=\mathrm{1}−{sin}^{\mathrm{2}} \left(\alpha−\beta\right) \\ $$$$\mathrm{2}\left({cos}\left(\gamma\right)−\frac{{cos}\left(\alpha−\beta\right)}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{{sin}^{\mathrm{2}} \left(\alpha−\beta\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2} \\ $$$$\mathrm{2}\left({cos}\left(\gamma\right)−\frac{{cos}\left(\alpha−\beta\right)}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{{sin}^{\mathrm{2}} \left(\alpha−\beta\right)}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${note}\:{that}\:{sin}^{\mathrm{2}} \:,\:\left(\:\:\right)^{\mathrm{2}} \:\geqslant\mathrm{0} \\ $$$$\therefore\:\mathrm{2}\underset{{cyc}} {\sum}{cos}^{\mathrm{2}} \left(\alpha\right)\geqslant\underset{{cyc}} {\sum}\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$
Commented by Mingma last updated on 18/Jun/23
Ver nice solution, sir!
Commented by York12 last updated on 04/Mar/24
∴ Σ_(cyc) (a^2 /(b^2 +c^2 ))≥(9/2)−3=(3/2) how is that useful
$$\therefore\:\underset{{cyc}} {\sum}\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\geqslant\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{3}=\frac{\mathrm{3}}{\mathrm{2}}\:\:\mathrm{how}\:\mathrm{is}\:\mathrm{that}\:\mathrm{useful}\: \\ $$$$ \\ $$

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