Step 1
Let the two positive numbers be $a$ and $b$
Given that the A.M. between $a$ and $b =A$ and G.M.$=G$
$\Rightarrow\:\large\frac{a+b}{2}$$=A$ and $\sqrt {ab}=G$
$\Rightarrow\:\large\frac{A}{G}=\frac{a+b}{2\sqrt {ab}}$
Step 2
Applying componendo and dividendo we get
$\large\frac{A+G}{A-G}=\frac{a+b+2\sqrt {ab}}{a+b-2\sqrt {ab}}$
We know that $a+b+2\sqrt {ab}=(\sqrt a+\sqrt b)^2$ and
$a+b-2\sqrt {ab}=(\sqrt a-\sqrt b)^2$
$\Rightarrow\:\large\frac{A+G}{A-G}=\frac{(\sqrt a+\sqrt b)^2}{(\sqrt a-\sqrt b)^2}$
Taking square root on both the sides we get
$\Rightarrow\:\sqrt{\large\frac{A+B}{A-B}}=\sqrt{\frac{(\sqrt a+\sqrt b)^2}{(\sqrt a-\sqrt b)^2}}=\frac{\sqrt a+\sqrt b}{\sqrt a-\sqrt b}$
Step 3
Again by applying componendo and dividendo we get
$\Rightarrow\:\large\frac{\sqrt {A+B}+\sqrt {A-B}}{\sqrt {A+B}-\sqrt {A-B}}=\frac{(\sqrt a+\sqrt b)+(\sqrt a-\sqrt b)}{(\sqrt a+\sqrt b)-(\sqrt a-\sqrt b)}$
$\Rightarrow\:\large\frac{\sqrt {A+B}+\sqrt {A-B}}{\sqrt {A+B}-\sqrt {A-B}}=\frac{2\sqrt a}{2\sqrt b}=\frac{\sqrt a}{\sqrt b}$
Step 4
Squaring on both the sides we get
$\Rightarrow\:\bigg(\large\frac{\sqrt {A+B}+\sqrt {A-B}}{\sqrt {A+B}-\sqrt {A-B}}\bigg)^2=\frac{ a}{ b}$
$\Rightarrow\:\large\frac{(\sqrt {A+B})^2+(\sqrt {A-B})^2+2\sqrt {A+B}.\sqrt {A-B}}{(\sqrt {A+B})^2+(\sqrt {A-B})^2-2\sqrt {A+B}.\sqrt {A-B}}=\frac{a}{b}$
$\Rightarrow\:\large\frac{(A+B)+(A-B)+2\sqrt {(A+B)(A-B)}}{(A+B)+(A-B)-2\sqrt {(A+B)(A-B)}}=\frac{a}{b}$
$\Rightarrow\:\large\frac{2A+2\sqrt {(A+b)(A-B)}}{2A-2\sqrt {(A+B)(A-B)}}=\frac{a}{b}$
$\Rightarrow\:\large\frac{A+\sqrt {(A+B)(A-B)}}{A-\sqrt {(A+B)(A-B)}}=\frac{a}{b}$
$\Rightarrow\:a=A+\sqrt {(A+B)(A-B)}$ and $b=A-\sqrt {(A+B)(A-b)}$
$\Rightarrow\:$ The two numbers $a$ and $b$ are given by
$A\pm \sqrt {(A+B)(A-B)}$
Hence proved.