The AM between two positive numbers a and b(a>b) is twice their GM.

To prove: Prove that a:b $=(2+\sqrt{3}):(2-\sqrt{3})$

Given: Arithmetic mean is twice of Geometric mean.

Formula used: (i) Arithmetic mean between a and $b=\frac{a+b}{2}$

(ii) Geometric mean between $a$ and $b=\sqrt{a b}$

AM = 2(GM)

$\frac{a+b}{2}=2(\sqrt{a b})$

$\Rightarrow a+b=4(\sqrt{a b})$

Squaring both side

$\Rightarrow(a+b)^{2}=16 a b \ldots(i)$

We know that $(a-b)^{2}=(a+b)^{2}-4 a b$

From eqn. (i)

$\Rightarrow(a-b)^{2}=16 a b-4 a b$

$\Rightarrow(a-b)^{2}=12 a b \ldots$ (ii)

Dividing eqn. (i) and (ii)

$\frac{(a+b)^{2}}{(a-b)^{2}}=\frac{16 a b}{12 a b}$

$\Rightarrow\left(\frac{a+b}{a-b}\right)^{2}=\frac{16}{12}$

Taking square root both side

$\Rightarrow \frac{a+b}{a-b}=\frac{4}{2 \sqrt{3}}$

$\Rightarrow \frac{a+b}{a-b}=\frac{2}{\sqrt{3}}$

Applying componendo and dividend

$\Rightarrow \frac{a+b+a-b}{a+b-a+b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$

$\Rightarrow \frac{2 a}{2 b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$

$\Rightarrow \frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$

Hence Proved