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Question Number 98022 by Ar Brandon last updated on 11/Jun/20

$$\mathcal{D}\mathrm{erive}\mathrm{the}\mathrm{relation}\mathrm{between}\mathrm{an}\mathcal{A}\mathrm{rithmetic}\mathcal{M}\mathrm{ean}$$$$\mathrm{and}\mathrm{a}\mathcal{G}\mathrm{eometric}\mathcal{M}\mathrm{ean}$$$$\sqrt[\mathrm{n}]{{\mathrm{x}}_1{\mathrm{x}}_2...{\mathrm{x}}_{\mathrm{n}}}⩽\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+\cdot \cdot \cdot +{\mathrm{x}}_{\mathrm{n}}}{\mathrm{n}}\mathrm{\forall }\mathrm{n}\in {\mathbb{N}}^{\ast },\mathrm{\forall }({\mathrm{x}}_1,{\mathrm{x}}_2,...{\mathrm{x}}_{\mathrm{n}})\in {\left({\mathbb{R}}_{+}^{\ast }\right)}^{\mathrm{n}}$$

Answered by Rio Michael last updated on 11/Jun/20

$$\mathrm{The}\mathrm{geometric}\mathrm{mean}G=\sqrt{ab}$$$$\mathrm{The}\mathrm{arithmetic}\mathrm{mean}A=\frac{a+b}{2}$$$$A-G>0$$$$A-G=\frac{a+b}{2}+\sqrt{ab}$$$$=\frac{a+b-2\sqrt{ab}}{2}$$$$=\frac{{(\sqrt{a}-\sqrt{b})}^2}{2}$$$$\mathrm{for}\mathrm{a}\mathrm{quadratic}{x}^2-2Ax+G^2=0\mathrm{where}{G}^2=\sqrt{ab}\mathrm{and}A=\frac{a+b}{2}$$$$\mathrm{has}\mathrm{roots}x=A\pm \sqrt{A^2-G^2}$$

Commented by Ar Brandon last updated on 11/Jun/20

$$\mathcal{T}\mathrm{hanks}\mathrm{for}\mathrm{your}\mathrm{time}.\mathcal{B}\mathrm{ut}\mathrm{actually},\mathrm{the}\mathrm{question}\mathrm{requires}$$$$\mathrm{us}\mathrm{to}\mathrm{derive}\mathrm{the}\mathrm{above}\mathrm{relation},\mathrm{instead}\mathrm{of}\mathrm{solving}\mathrm{for}\mathrm{x}.$$

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