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Question Number 197706 by universe last updated on 26/Sep/23

Answered by witcher3 last updated on 27/Sep/23

$${\left(\frac{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}{2}\right)}^2⩽\frac{\mathrm{a}+\mathrm{b}}{2}\iff \frac{\mathrm{a}+\mathrm{b}}{4}⩾\frac{\sqrt{\mathrm{ab}}}{2}.\mathrm{Am}-\mathrm{GM}$$$$\Rightarrow \frac{1}{\sqrt{1+{\mathrm{x}}^2}}+\frac{1}{\sqrt{1+{\mathrm{y}}^2}}⩽2\sqrt{\frac{1}{2}(\frac{1}{1+{\mathrm{x}}^2}+\frac{1}{1+{\mathrm{y}}^2}})$$$$\frac{1}{1+{\mathrm{x}}^2}+\frac{1}{1+{\mathrm{y}}^2}⩽\frac{2}{1+\mathrm{xy}}......?$$$$\mathrm{we}\mathrm{have}\mathrm{by}\mathrm{symetrie}\mathrm{x}>\mathrm{y}$$$$\frac{1}{(1+{\mathrm{x}}^2)}+\frac{1}{1+{\mathrm{y}}^2}-\frac{2}{1+\mathrm{xy}}⩽0\iff$$$$(1+{\mathrm{y}}^2)(1+\mathrm{xy})+(1+{\mathrm{x}}^2)(1+\mathrm{xy})-2(1+{\mathrm{x}}^2)(1+{\mathrm{y}}^2)⩽00$$$$(1+\mathrm{xy})(2+{\mathrm{x}}^2+{\mathrm{y}}^2)$$$$-{\mathrm{x}}^2-{\mathrm{y}}^2+2\mathrm{xy}+\mathrm{xy}({\mathrm{x}}^2+{\mathrm{y}}^2)-{2\mathrm{x}}^2{\mathrm{y}}^2$$$$=({\mathrm{x}}^2+{\mathrm{y}}^2)(\mathrm{xy}-1)-2\mathbf{xy}(\mathbf{xy}-1)$$$$=(\mathbf{xy}-1){(\mathbf{x}-\mathbf{y})}^2⩽0,\mathrm{xy}⩽1$$$$\Rightarrow \frac{1}{2}(\frac{1}{1+{\mathrm{x}}^2}+\frac{1}{1+{\mathrm{y}}^2})⩽\frac{1}{1+\mathrm{xy}}$$$$\Rightarrow \frac{1}{\sqrt{1+{\mathrm{x}}^2}}+\frac{1}{\sqrt{1+{\mathrm{y}}^2}}⩽2\sqrt{\frac{1}{2}(\frac{1}{1+{\mathrm{x}}^2}+\frac{1}{1+{\mathrm{y}}^2})}⩽2.\frac{1}{\sqrt{1+\mathrm{xy}}}$$$$$$$$$$$$$$$$$$$$$$$$$$

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