Let the two numbers be $a$ and $b$.
Then, $A=\frac{a+b}{2}$
$\Rightarrow 2 A=a+b$
and $G_1, G_2$ are geometric mean between $a$ and $b$, then a, $G_1, G_2, b$ are in GP.
Let $r$ be the common ratio.
$$
\begin{aligned}
&\therefore b=a r^{4-1} \quad\left[\because a_n=a r^{n-1}\right] \\
&\Rightarrow b=a r^3 \Rightarrow \frac{b}{a}=r^3 \Rightarrow r=\left(\frac{b}{a}\right)^{\frac{1}{3}}
\end{aligned}
$$
Now, $G_1=a r=a\left(\frac{b}{a}\right)^{\frac{1}{3}}$
and $G_2=\operatorname{ar}^2=a\left(\frac{b}{a}\right)^{\frac{2}{3}}$
$\mathrm{RHS}=\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=\frac{\left[a\left(\frac{b}{a}\right)^{1 / 3}\right]^2}{a\left(\frac{b}{a}\right)^{2 / 3}}+\frac{\left[a\left(\frac{b}{a}\right)^{2 / 3}\right]^2}{a\left(\frac{b}{a}\right)^{1 / 3}}$
$=\frac{a^2\left(\frac{b}{a}\right)^{2 / 3}}{a\left(\frac{b}{a}\right)^{2 / 3}}+\frac{a^2\left(\frac{b}{a}\right)^{4 / 3}}{a\left(\frac{b}{a}\right)^{1 / 3}}=a+a\left(\frac{b}{a}\right)=a+b=2 A$
$=$ LHS $\quad$ [using Eq. (i)]