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Question Number 193138 by Mastermind last updated on 04/Jun/23

$$\mathrm{Show}\mathrm{that}\mathrm{for}\mathrm{all}\mathrm{a},\mathrm{b}\in \mathbb{R}$$$$\mathrm{i})\mathrm{ab}⩽\frac{1}{2}({\mathrm{a}}^2+{\mathrm{b}}^2)$$$$\mathrm{ii}){\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^2⩽({\mathrm{a}}^2+{\mathrm{b}}^2)$$$$\mathrm{iii})\sqrt{\mathrm{ab}}⩽\frac{1}{2}(\mathrm{a}+\mathrm{b}),\mathrm{for}\mathrm{a},\mathrm{b}⩾0\mathrm{such}\mathrm{that}$$$$\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{have}\mathrm{square}\mathrm{roots}.$$$$$$$$\mathrm{Help}$$

Answered by Subhi last updated on 04/Jun/23

$$\left(i\right){(a-b)}^2⩾0$$$$a^2+b^2-2ab⩾0$$$$ab⩽\frac{1}{2}(a^2+b^2)$$$$\left(ii\right){(a+b)}^2=a^2+b^2+2ab$$$$notethat2ab⩽a^2+b^2$$$$a^2+b^2+2ab⩽2(a^2+b^2)$$$$\frac{{(a+b)}^2}{2}⩽a^2+b^2$$$$\left(iii\right){(\sqrt{a}-\sqrt{b})}^2⩾0$$$$a+b-2\sqrt{ab}⩾0$$$$\sqrt{ab}⩽\frac{1}{2}(a+b)$$

Answered by aba last updated on 04/Jun/23

$$\mathrm{i}){(\mathrm{a}-\mathrm{b})}^2⩾0\Rightarrow \frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}⩾\mathrm{ab}$$$$\mathrm{ii})\mathrm{ab}=\frac{1}{2}({(\mathrm{a}+\mathrm{b})}^2-({\mathrm{a}}^2-{\mathrm{b}}^2))⩽\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}\Rightarrow \frac{({\mathrm{a}}^2+{\mathrm{b}}^2)}{2}⩽{\mathrm{a}}^2+{\mathrm{b}}^2$$$$\mathrm{iii}){(\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}})}^2⩾0\Rightarrow \mathrm{ab}⩽\frac{\mathrm{a}+\mathrm{b}}{2}$$$$$$

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