Question

The ratio of the AM and GM of two positive numbers a and b is m : n, show that $a:b=[m+\sqrt{(m^2-n^2)}]:[m-\sqrt{(m^2-n^2)}].$

Answer 1

Given, $\frac{AM}{GM}=\frac{m}{n}$

$\Rightarrow \frac{a+b}{2\sqrt{ab}}=\frac{m}{n}$

$\Rightarrow \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{m+n}{m-n}$

[applying componendo and dividendo rule]

$\Rightarrow \frac{(\sqrt{a}+\sqrt{b}{)}^2}{(\sqrt{a}-\sqrt{b}{)}^2}=\frac{m+n}{m-n}$

$\Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{m+n}}{\sqrt{m-n}}$

$\Rightarrow \frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$

[again applying componendo and dividendo rule]

$\Rightarrow \frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$

On squaring both sides, we get

$\Rightarrow \frac{a}{b}=\frac{(m+n)+(m-n)+2\sqrt{(m+n)(m-n)}}{(m+n)+(m-n)-2\sqrt{(m+n)(m-n)}}$

$\Rightarrow \frac{a}{b}=\frac{2m+2\sqrt{m^2-n^2}}{2m-2\sqrt{m^2-n^2}}$

$\therefore \frac{a}{b}=\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}$

Hence proved.