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

$$x;y\in {\mathbb{R}}^{+};x^2+y^2=2$$$$proof:3-xy⩾(x+y)\sqrt{xy}+{(x-y)}^2⩾2xy$$

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

$${we}^{\prime }llusea+b⩾2\sqrt{ab}$$$$$$$$x^2+y^2=2$$$$\Rightarrow x=\sqrt{2}\cos \theta >0$$$$\Rightarrow y=\sqrt{2}\sin \theta >0$$$$$$$$(x+y)\sqrt{xy}+{(x-y)}^2$$$$=\sqrt{2}(\cos \theta +\sin \theta )\sqrt{2\cos \theta \sin \theta }+2{(\cos \theta -\sin \theta )}^2$$$$=(\cos \theta +\sin \theta )2\sqrt{\cos \theta \sin \theta }+2-4\cos \theta \sin \theta$$$$⩽(\cos \theta +\sin \theta )(\cos \theta +\sin \theta )+2-4\cos \theta \sin \theta$$$$=1+2\cos \theta \sin \theta +2-4\cos \theta \sin \theta$$$$=3-2\cos \theta \sin \theta$$$$=3-xy$$$$$$$$(x+y)\sqrt{xy}+{(x-y)}^2$$$$=(\cos \theta +\sin \theta )2\sqrt{\cos \theta \sin \theta }+2-4\cos \theta \sin \theta$$$$⩾2\sqrt{\cos \theta \sin \theta }2\sqrt{\cos \theta \sin \theta }+2-4\cos \theta \sin \theta$$$$=4\cos \theta \sin \theta +2-4\cos \theta \sin \theta$$$$=2$$$$⩾2\sin 2\theta =2\times 2\cos \theta \sin \theta$$$$=2xy$$

Commented by mathdanisur last updated on 05/May/21

$$coolthankyousir$$

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