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If the arithmetic, geometric and harmonic means between two distinct positive real numbers be $A, G$ and H respectively, then the relation between them is
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The correct answer is:
$A \gt G \gt H$
$\mathrm{A}=\frac{(\mathrm{a}+\mathrm{b})}{2}, G=\sqrt{a b}$ and $H=\frac{2 a b}{(a+b)}$.
$A-G=\frac{(a+b)}{2}-\sqrt{a b}$
$=\frac{(a+b-2 \sqrt{a b})}{2}$
$=\frac{(\sqrt{a}-\sqrt{b})^2}{2} \geq 0$
So $A \geq G$
Step 2: Relation between $G ~\&~ H$
$G-H=\sqrt{a b}-\frac{2 a b}{(a+b)}$
$=\frac{\sqrt{a b}[a+b-2 \sqrt{(a b)]}}{(a+b)}$
$=$ $\frac{\sqrt{(a b)}(\sqrt{a}-\sqrt{b})^2}{(a+b)}$ $\geq 0$
$G \geq H$
Thus $\mathrm{A} \geqslant \mathrm{G} \geq \mathrm{H}$.
$A-G=\frac{(a+b)}{2}-\sqrt{a b}$
$=\frac{(a+b-2 \sqrt{a b})}{2}$
$=\frac{(\sqrt{a}-\sqrt{b})^2}{2} \geq 0$
So $A \geq G$
Step 2: Relation between $G ~\&~ H$
$G-H=\sqrt{a b}-\frac{2 a b}{(a+b)}$
$=\frac{\sqrt{a b}[a+b-2 \sqrt{(a b)]}}{(a+b)}$
$=$ $\frac{\sqrt{(a b)}(\sqrt{a}-\sqrt{b})^2}{(a+b)}$ $\geq 0$
$G \geq H$
Thus $\mathrm{A} \geqslant \mathrm{G} \geq \mathrm{H}$.
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