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Question Number 140129 by mathdanisur last updated on 04/May/21

x;y∈R^+ ; x^2 +y^2 =2 proof: 3−xy≥(x+y)(√(xy))+(x−y)^2 ≥2xy

$${x};{y}\in\mathbb{R}^{+} \:;\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$${proof}:\:\mathrm{3}−{xy}\geqslant\left({x}+{y}\right)\sqrt{{xy}}+\left({x}−{y}\right)^{\mathrm{2}} \geqslant\mathrm{2}{xy} \\ $$

Answered by mr W last updated on 05/May/21

we′ll use a+b≥2(√(ab)) x^2 +y^2 =2 ⇒x=(√2)cos θ>0 ⇒y=(√2)sin θ>0 (x+y)(√(xy))+(x−y)^2 =(√2)(cos θ+sin θ)(√(2cos θsin θ))+2(cos θ−sin θ)^2 =(cos θ+sin θ)2(√(cos θsin θ))+2−4cos θsin θ ≤(cos θ+sin θ)(cos θ+sin θ)+2−4cos θsin θ =1+2cos θsin θ+2−4cos θsin θ =3−2cos θsin θ =3−xy (x+y)(√(xy))+(x−y)^2 =(cos θ+sin θ)2(√(cos θsin θ))+2−4cos θsin θ ≥2(√(cos θsin θ))2(√(cos θsin θ))+2−4cos θsin θ =4cos θsin θ+2−4cos θsin θ =2 ≥2 sin 2θ=2×2cos θsin θ =2xy

$${we}'{ll}\:{use}\:{a}+{b}\geqslant\mathrm{2}\sqrt{{ab}} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{2}}\mathrm{cos}\:\theta>\mathrm{0} \\ $$$$\Rightarrow{y}=\sqrt{\mathrm{2}}\mathrm{sin}\:\theta>\mathrm{0} \\ $$$$ \\ $$$$\left({x}+{y}\right)\sqrt{{xy}}+\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$=\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\sqrt{\mathrm{2cos}\:\theta\mathrm{sin}\:\theta}+\mathrm{2}\left(\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\mathrm{2}\sqrt{\mathrm{cos}\:\theta\mathrm{sin}\:\theta}+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$\leqslant\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{1}+\mathrm{2cos}\:\theta\mathrm{sin}\:\theta+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{3}−\mathrm{2cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{3}−{xy} \\ $$$$ \\ $$$$\left({x}+{y}\right)\sqrt{{xy}}+\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\mathrm{2}\sqrt{\mathrm{cos}\:\theta\mathrm{sin}\:\theta}+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$\geqslant\mathrm{2}\sqrt{\mathrm{cos}\:\theta\mathrm{sin}\:\theta}\mathrm{2}\sqrt{\mathrm{cos}\:\theta\mathrm{sin}\:\theta}+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{4cos}\:\theta\mathrm{sin}\:\theta+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{2} \\ $$$$\geqslant\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2}×\mathrm{2cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{2}{xy} \\ $$

Commented by mathdanisur last updated on 05/May/21

cool thankyou sir

$${cool}\:{thankyou}\:{sir} \\ $$

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