Question Number 140129 by mathdanisur last updated on 04/May/21
$${x};{y}\in\mathbb{R}^{+} \:;\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$${proof}:\:\mathrm{3}−{xy}\geqslant\left({x}+{y}\right)\sqrt{{xy}}+\left({x}−{y}\right)^{\mathrm{2}} \geqslant\mathrm{2}{xy} \\ $$
Answered by mr W last updated on 05/May/21
$${we}'{ll}\:{use}\:{a}+{b}\geqslant\mathrm{2}\sqrt{{ab}} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{2}}\mathrm{cos}\:\theta>\mathrm{0} \\ $$$$\Rightarrow{y}=\sqrt{\mathrm{2}}\mathrm{sin}\:\theta>\mathrm{0} \\ $$$$ \\ $$$$\left({x}+{y}\right)\sqrt{{xy}}+\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$=\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\sqrt{\mathrm{2cos}\:\theta\mathrm{sin}\:\theta}+\mathrm{2}\left(\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\mathrm{2}\sqrt{\mathrm{cos}\:\theta\mathrm{sin}\:\theta}+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$\leqslant\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{1}+\mathrm{2cos}\:\theta\mathrm{sin}\:\theta+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{3}−\mathrm{2cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{3}−{xy} \\ $$$$ \\ $$$$\left({x}+{y}\right)\sqrt{{xy}}+\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\mathrm{2}\sqrt{\mathrm{cos}\:\theta\mathrm{sin}\:\theta}+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$\geqslant\mathrm{2}\sqrt{\mathrm{cos}\:\theta\mathrm{sin}\:\theta}\mathrm{2}\sqrt{\mathrm{cos}\:\theta\mathrm{sin}\:\theta}+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{4cos}\:\theta\mathrm{sin}\:\theta+\mathrm{2}−\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{2} \\ $$$$\geqslant\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2}×\mathrm{2cos}\:\theta\mathrm{sin}\:\theta \\ $$$$=\mathrm{2}{xy} \\ $$
Commented by mathdanisur last updated on 05/May/21
$${cool}\:{thankyou}\:{sir} \\ $$