A.M. = 2 × G.M. [ \dfrac{a+b}{2}=2\sqrt{ab}\implies a+b=4\sqrt{ab} ] [ \sqrt{a}+\sqrt{b}\ \text{or divide by }\sqrt{ab}: ] [ \dfrac{a}{b}+1=4\sqrt{\dfrac{a}{b}}\quad\text{let }k=\sqrt{\dfrac{a}{b}} ] [ k^2-4k+1=0\implies k=2\pm\sqrt{3} ] [ \dfrac{a}{b}=(2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3} ] Better: (\dfrac{a}{b}=\dfrac{(2+\sqrt{3})^2}{1}) vs ratio form requested:

[ a:b = (2+\sqrt{3})^2 : 1 ? ]

From (k=2+\sqrt{3}) (taking (a>b)): [ \sqrt{\dfrac{a}{b}}=2+\sqrt{3}\implies\dfrac{a}{b}=(2+\sqrt{3})^2=7+4\sqrt{3} ]

But the required form is (a:b=2+\sqrt{3}:2-\sqrt{3}): [ \dfrac{2+\sqrt{3}}{2-\sqrt{3}}\cdot\dfrac{2+\sqrt{3}}{2+\sqrt{3}}=\dfrac{(2+\sqrt{3})^2}{4-3}=7+4\sqrt{3} ] Yes matches.

Proved.