If $A, A1, A2, A3$ are the areas of incircle and excircle of a triangle respectively, then prove that $\frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}}=\frac{1}{\sqrt{A}}$

Given $A, A1, A2, A3$ are the areas of incircle and Excircle of a triangle respectively, Then $A=\pi r^2$

$A_1=\pi r_1^2,A_2=\pi r_2^2,A_3=\pi r_3^2$

Then $\frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}}=\frac{1}{\sqrt{\pi r_1^2}}+\frac{1}{\sqrt{\pi r_2^2}}+\frac{1}{\sqrt{\pi r_3^2}}$

$=\frac{1}{\sqrt{\pi }{r}_1}+\frac{1}{\sqrt{\pi }{r}_2}+\frac{1}{\sqrt{\pi }{r}_3}=\frac{1}{\sqrt{\pi }}\left[\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\right]$

We know that, $\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{r}$

$=\frac{1}{\sqrt{\pi }}\left(\frac{1}{r}\right)=\frac{1}{\sqrt{\pi r^2}}=\frac{1}{\sqrt{A}}$

$\therefore \frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}}=\frac{1}{\sqrt{A}}$