The ratio of the A.M and G.M of two positive numbers $a$ and $b$ , is $m : n$ . Show that $a : b = (m + \sqrt{m^{2} - n^{2}}) : (m - \sqrt{m^{2} - n^{2}})$

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Solution

Positive number $a, b$
A.M $= \frac{a + b}{2}$ $G M = \sqrt{a b}$
$m : n = \frac{a + b}{2 \sqrt{a b}}$
$R . H . S = \frac{(m + \sqrt{m^{2} - n^{2}})}{(m - \sqrt{m^{2} - n^{2}})}$
Divide by n in both numerator and denominator.
$= \frac{(\frac{m}{n} + \sqrt{\frac{m^{2}}{n^{2}} - 1})}{(\frac{m}{n} - \sqrt{\frac{m^{2}}{n^{2}} - 1})}$
$= \frac{(\frac{a + b}{2 \sqrt{a b}} + \sqrt{\frac{a^{2} + b^{2} + 2 a b}{4 a b} - 1})}{(\frac{a + b}{2 \sqrt{a b}} - \sqrt{\frac{a^{2} + b^{2} + 2 a b}{4 a b} - 1})}$
$= \frac{(\frac{a + b}{2 \sqrt{a b}} + \sqrt{\frac{a^{2} + b^{2} - 2 a b}{4 a b}})}{(\frac{a + b}{2 \sqrt{a b}} - \sqrt{\frac{a^{2} + b^{2} - 2 a b}{4 a b}})}$
$= \frac{a + b + (a - b)}{2 \sqrt{a b}}$
$\frac{2 a}{2 b} = \frac{a}{b} = L H S$
$L H S = R H S$

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