Question-197706

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

Answered by witcher3 last updated on 27/Sep/23

((((√a)+(√b))/2))^2 ≤((a+b)/2)⇔((a+b)/4)≥((√(ab))/2).Am−GM ⇒(1/( (√(1+x^2 ))))+(1/( (√(1+y^2 ))))≤2(√((1/2)((1/(1+x^2 ))+(1/(1+y^2 ))))) (1/(1+x^2 ))+(1/(1+y^2 ))≤(2/(1+xy))......? we have by symetrie x>y (1/((1+x^2 )))+(1/(1+y^2 ))−(2/(1+xy))≤0⇔ (1+y^2 )(1+xy)+(1+x^2 )(1+xy)−2(1+x^2 )(1+y^2 )≤00 (1+xy)(2+x^2 +y^2 ) −x^2 −y^2 +2xy+xy(x^2 +y^2 )−2x^2 y^2 =(x^2 +y^2 )(xy−1)−2xy(xy−1) =(xy−1)(x−y)^2 ≤0, xy≤1 ⇒(1/2)((1/(1+x^2 ))+(1/(1+y^2 )))≤(1/(1+xy)) ⇒(1/( (√(1+x^2 ))))+(1/( (√(1+y^2 ))))≤2(√((1/2)((1/(1+x^2 ))+(1/(1+y^2 )))))≤2.(1/( (√(1+xy))))

$$\left(\frac{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\Leftrightarrow\frac{\mathrm{a}+\mathrm{b}}{\mathrm{4}}\geqslant\frac{\sqrt{\mathrm{ab}}}{\mathrm{2}}.\mathrm{Am}−\mathrm{GM} \\ $$$$\left.\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }}\leqslant\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\right.}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\leqslant\frac{\mathrm{2}}{\mathrm{1}+\mathrm{xy}}……? \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{by}\:\mathrm{symetrie}\:\:\mathrm{x}>\mathrm{y} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }−\frac{\mathrm{2}}{\mathrm{1}+\mathrm{xy}}\leqslant\mathrm{0}\Leftrightarrow \\ $$$$\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{xy}\right)+\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{xy}\right)−\mathrm{2}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)\leqslant\mathrm{00} \\ $$$$\left(\mathrm{1}+\mathrm{xy}\right)\left(\mathrm{2}+\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right) \\ $$$$−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{xy}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)−\mathrm{2x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \\ $$$$=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{xy}−\mathrm{1}\right)−\mathrm{2}\boldsymbol{\mathrm{xy}}\left(\boldsymbol{\mathrm{xy}}−\mathrm{1}\right) \\ $$$$=\left(\boldsymbol{\mathrm{xy}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}\right)^{\mathrm{2}} \leqslant\mathrm{0},\:\mathrm{xy}\leqslant\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\right)\leqslant\frac{\mathrm{1}}{\mathrm{1}+\mathrm{xy}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }}\leqslant\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\right)}\leqslant\mathrm{2}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{xy}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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