Question Number 155594 by mathdanisur last updated on 02/Oct/21
Answered by ghimisi last updated on 02/Oct/21
$$\Leftrightarrow\Sigma\left(\sqrt{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){xy}\right)\geqslant\mathrm{0} \\ $$$$\sqrt{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){xy}\geqslant\mathrm{0}\Leftrightarrow \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){xy}\geqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\sqrt{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }\Leftrightarrow \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){xy}\geqslant\frac{\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }}\Leftrightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }\geqslant\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){xy}\:\left(\bullet\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \geqslant\mathrm{2}{xy}\:\:\:\left(\mathrm{1}\right) \\ $$$$\sqrt{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }\geqslant\sqrt{\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} }={xy}\sqrt{\mathrm{2}}\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow\left(\bullet\right) \\ $$$$ \\ $$
Commented by mathdanisur last updated on 02/Oct/21
$$\mathrm{Perfect}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by amin96 last updated on 02/Oct/21
$${great} \\ $$