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Question Number 47476 by hassentimol last updated on 10/Nov/18

$$\mathrm{How}\mathrm{may}\mathrm{I}\mathrm{prove}\mathrm{the}\mathrm{following}\mathrm{theorem}?$$$$$$$$\frac{\mathit{a}+\mathit{b}}{2}⩾\sqrt{\mathit{a}\mathit{b}}$$$$$$$$\mathrm{Thank}\mathrm{you}$$

Commented by prakash jain last updated on 11/Nov/18

$$\mathrm{This}\mathrm{is}\mathrm{true}\mathrm{only}\mathrm{for}a,b⩾0$$

Answered by Joel578 last updated on 11/Nov/18

$$\mathrm{Assume}a,b\in \mathbb{R},\mathrm{then}\sqrt{a},\sqrt{b}\in \mathbb{R}$$$$\mathrm{Observe}\mathrm{that}\mathrm{for}\mathrm{all}\sqrt{a},\sqrt{b}\in \mathbb{R}$$$${(\sqrt{a}-\sqrt{b})}^2⩾0$$$$\iff a-2\sqrt{ab}+b⩾0$$$$\iff a+b⩾2\sqrt{ab}$$$$\iff \frac{a+b}{2}⩾\sqrt{ab}$$$$\mathrm{Hence},\mathrm{proved}$$

Commented by hassentimol last updated on 11/Nov/18

$$$$$$\mathrm{Thank}\mathrm{you}\mathrm{sir}!$$$$\mathrm{It}\mathrm{is}\mathrm{also}\mathrm{very}\mathrm{helpful}!$$

Answered by .... last updated on 10/Nov/18

$$\mathrm{since}{(\mathrm{a}-\mathrm{b})}^2⩾0$$$$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}⩾0$$$$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}+4\mathrm{ab}⩾0+4\mathrm{ab}$$$$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}⩾4\mathrm{ab}$$$$\Rightarrow {(\mathrm{a}+\mathrm{b})}^2⩾4\mathrm{ab}$$$$\Rightarrow (\mathrm{a}+\mathrm{b})⩾2\sqrt{\mathrm{ab}}$$$$\Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2}⩾\sqrt{\mathrm{ab}}$$$$$$$$$$$$\frac{\frac{}{}}{}$$

Commented by hassentimol last updated on 11/Nov/18

$$$$$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.$$$$\mathrm{It}\mathrm{is}\mathrm{very}\mathrm{helpful}!$$

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