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Let $a$ and $b$ be two positive real numbers. Suppose
$A_{1}, A_{2}$ are two arithmetic means; $G_{1}, G_{2}$
are two geometric means and $H_{1}, H_{2}$ are two harmonic means between
a and $b,$ then $\frac{A_{1} + A_{2}}{H_{1} + H_{2}}$ is equal to

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$\frac{2 (a^{2} + b^{2}) + 5 a b}{9 a b}$

$\frac{a^{2} + b^{2}}{9 a b} + 5$

$\frac{a^{2} + b^{2} + 5 (a + b)}{9 a b}$

$\frac{a^{2} + b^{2} + 7 (a + b)}{3 (a + b) a b}$

answer is A.

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Detailed Solution

We have

$A_{1} = a + \frac{1}{3} (b - a) = \frac{2 a + b}{3}$

and $A_{2} = \frac{a + 2 b}{3}$

$∴ A_{1} + A_{2} = a + b$

Also $\frac{1}{H_{1}} + \frac{1}{H_{2}} = \frac{1}{a} + \frac{1}{b}$

and $\frac{1}{H_{2}} = \frac{1}{3} (\frac{2 a + b}{a b})$

$\Rightarrow H_{1} + H_{2} = \frac{a + b}{a b} H_{1} H_{2}$

Therefore,

$\begin{array}{l} \frac{A_{1} + A_{2}}{H_{1} + H_{2}} = \frac{a b}{H_{1} H_{2}} = \frac{a b}{9} \frac{(a + 2 b) (2 a + b)}{(a b)^{2}} \\ = \frac{2 (a^{2} + b^{2}) + 5 a b}{9 a b} \end{array}$

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