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Question Number 151448 by mathdanisur last updated on 21/Aug/21

$$\mathrm{Prove}\mathrm{that}:$$$$\mathbf{A}\mathrm{rtimetric}\mathrm{mean}⩾\mathbf{G}\mathrm{eometric}\mathrm{mean}$$$$\frac{\mathrm{a}+\mathrm{b}}{2}⩾\sqrt{\mathrm{ab}}$$

Commented by puissant last updated on 21/Aug/21

$$\mathrm{\forall }(a,b)\in {\mathbb{R}}^2,$$$${(\sqrt{a}-\sqrt{b})}^2⩾0$$$$\Rightarrow a-2\sqrt{ab}+b⩾0$$$$\Rightarrow 2\sqrt{ab}⩽a+b$$$$\Rightarrow \sqrt{ab}⩽\frac{a+b}{2}..$$

Commented by mathdanisur last updated on 21/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Ser}$$

Answered by nimnim last updated on 21/Aug/21

$$\mathrm{This}\mathrm{inequality}\mathrm{hold}\mathrm{when}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}\mathrm{positive}.$$$$\mathrm{Since}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}\mathrm{positive},$$$$\sqrt{\mathrm{a}}=\mathrm{x}\mathrm{and}\sqrt{\mathrm{b}}=\mathrm{y}\left(\mathrm{say}\right)$$$$\mathrm{then}{(\mathrm{x}-\mathrm{y})}^2⩾0$$$$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{xy}⩾0$$$$\Rightarrow \frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{2}⩾\mathrm{xy}$$$$\Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2}⩾\sqrt{\mathrm{ab}}★$$

Commented by mathdanisur last updated on 21/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Ser}$$

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