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Question Number 175396 by Shrinava last updated on 29/Aug/22

Answered by mahdipoor last updated on 29/Aug/22

$$x+y^2\sqrt{\frac{2}{x^2+y^2}}⩾y+\sqrt{\frac{x^2+y^2}{2}}\iff$$$$(x+y^2\sqrt{\frac{2}{x^2+y^2}})\left(\sqrt{\frac{x^2+y^2}{2}}\right)⩾(y+\sqrt{\frac{x^2+y^2}{2}})\left(\sqrt{\frac{x^2+y^2}{2}}\right)\iff$$$$x\sqrt{\frac{x^2+y^2}{2}}+y^2⩾y\sqrt{\frac{x^2+y^2}{2}}+\frac{x^2+y^2}{2}\iff$$$$(x-y)\sqrt{\frac{x^2+y^2}{2}}⩾\frac{x^2-y^2}{2}=\frac{(x-y)(x+y)}{2}\stackrel{\mathrm{I}}{\iff }$$$$\sqrt{\frac{x^2+y^2}{2}⩾}\frac{x+y}{2}\stackrel{\mathrm{II}}{\iff }\frac{x^2+y^2}{2}⩾\frac{x^2+y^2+2xy}{4}\iff$$$$\frac{x^2+y^2-2xy}{4}=\frac{{(x-y)}^2}{4}⩾0$$$$\mathrm{I}:x-y⩾0orx⩾y$$$$\mathrm{II}:x+y⩾0$$

Answered by MJS_new last updated on 29/Aug/22

$$\mathrm{let}y=px\mathrm{with}p>0$$$$x\frac{\sqrt{2}{p}^2+\sqrt{p^2+1}}{\sqrt{p^2+1}}⩾x\frac{\sqrt{2}p+\sqrt{p^2+1}}{\sqrt{2}}$$$$\frac{\sqrt{2}{p}^2+\sqrt{p^2+1}}{\sqrt{p^2+1}}⩾\frac{\sqrt{2}p+\sqrt{p^2+1}}{\sqrt{2}}$$$$2p^2+\sqrt{2(p^2+1)}⩾p^2+1+p\sqrt{2(p^2+1)}$$$$(p-1)(p+1-\sqrt{2(p^2+1)})⩾0$$$$\mathrm{only}\mathrm{true}\mathrm{for}0<p⩽1$$$$\Rightarrow \mathrm{only}\mathrm{true}\mathrm{for}0<y⩽x$$

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