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If $a$ and $b$ are two different positive real numbers, then which of the following relations is true
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The correct answer is:
$2 \sqrt{a b} \lt (a+b)$
We know that $A \gt G \gt H$
Where $A$ is arithmetic mean, $G$ is geometric mean and $H$ is harmonic mean, then $A \gt G$
$\Rightarrow \frac{a+b}{2} \gt \sqrt{a b}$ or $(a+b) \gt 2 \sqrt{a b}$
Where $A$ is arithmetic mean, $G$ is geometric mean and $H$ is harmonic mean, then $A \gt G$
$\Rightarrow \frac{a+b}{2} \gt \sqrt{a b}$ or $(a+b) \gt 2 \sqrt{a b}$
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