<s>[INST] Triangle Area problem I've been trying to solve the following:
Let $ABC$ be a triangle with sides $a, b $ and $ c$, inradius $r$ and exradii $r_a, r_b$ and $r_c$. If $A'B'C'$ is another triangle with sides $\sqrt{a}, \sqrt{b}$ and $\sqrt{c}$ show that $Area(A'B'C')=\frac {\sqrt{r(r_a+r_b+r_c)}} {2}$. 
I tried to combine various formulas and theorems involving exradii and inradii but it got me nowhere so far.. maybe there's a better way to approach the problem! Any advice or help will be appreciated!!
 [/INST] Using
$$\text{area($\triangle{ABC}$)}=\frac{a+b+c}{2}r=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4}$$
$$\Rightarrow \frac{a+b+c}{2}r\cdot\frac{1}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=\frac 14$$
and
$$\text{area($\triangle{ABC}$)}=\frac{a+b+c}{2}r=\frac{-a+b+c}{2}r_a=\frac{a-b+c}{2}r_b=\frac{a+b-c}{2}r_c$$$$\Rightarrow rr_a=\frac{a+b+c}{-a+b+c}r^2,\quad rr_b=\frac{a+b+c}{a-b+c}r^2,\quad rr_c=\frac{a+b+c}{a+b-c}r^2$$we have
$$\begin{align}\text{area($\triangle{A'B'C'}$)}&=\frac{\sqrt{(\sqrt a+\sqrt b+\sqrt c)(-\sqrt a+\sqrt b+\sqrt c)(\sqrt a-\sqrt b+\sqrt c)(\sqrt a+\sqrt b-\sqrt c)}}{4}\\&=\frac{\sqrt{2ab+2bc+2ca-a^2-b^2-c^2}}{4}\\&=\frac{a+b+c}{2}r\cdot \frac{\sqrt{2ab+2bc+2ca-a^2-b^2-c^2}}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}\\&=\frac r2\sqrt{\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{(-a+b+c)(a-b+c)(a+b-c)}}\\&=\frac r2\sqrt{(a+b+c)\left(\frac{1}{-a+b+c}+\frac{1}{a-b+c}+\frac{1}{a+b-c}\right)}\\&=\frac 12\sqrt{\frac{a+b+c}{-a+b+c}r^2+\frac{a+b+c}{a-b+c}r^2+\frac{a+b+c}{a+b-c}r^2}\\&=\frac{\sqrt{rr_a+rr_b+rr_c}}{2}\end{align}$$
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