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!!

Brute force method:
Firstly, Heron's formula can be simplified in this form:
The area $\Delta$ of a triangle with sides $a,b,c$ is given by $16\Delta^2 $ $= (a + b + c)(b + c - a) (c + a - b)(a + b - c) \\= 2(a^2 b^2 + b^2 c^2 + c^2 a^2) - (a^4 + b^4 + c^4)$
Therefore area of triangle with sides as $\sqrt a, \sqrt b, \sqrt c$ is $\frac 1 4 \sqrt{2 (ab + bc + ca) - (a^2 + b^2 + c^2)}$.
Also we have $r_a + r_b + r_c = 4R + r$. (Steiner's formula)
So it is enough to show that
$ \frac{\sqrt{r(r_a+r_b+r_c)}}{2} = \frac{1}4\sqrt{2 (ab + bc + ca) - (a^2 + b^2 + c^2)}$
which is equivalent to
$4r (4R + r) = 2 (ab + bc + ca) - (a^2 + b^2 + c^2)$.
Also $\Delta = \frac 12 (a + b + c)\cdot r = \frac{abc}{4R}$, so
$4r (4R + r) = 2 (ab + bc + ca) - (a^2 + b^2 + c^2)$
$\Leftrightarrow 4 \frac{2 \Delta}{a + b + c} \left(\frac{abc}{\Delta} + \frac{2 \Delta}{a + b + c} \right) = 2 (ab + bc + ca) - (a^2 + b^2 + c^2)$
(Note that here $\Delta$ is the area of the triangle with sides $a,b,c$)
$\Leftrightarrow 8 (abc (a + b + c) + 2 \Delta^2) = (a + b + c)^2(2 (ab + bc + ca) - (a^2 + b^2 + c^2))$
$\Leftrightarrow 8 abc (a + b + c) + (a + b + c)(b + c - a) (c + a - b)( a + b - c) = (a + b + c)^2(2(ab + bc + ca) - (a^2 + b^2 + c^2))$
$\Leftrightarrow 8 abc + (b + c - a)(c + a - b)(a + b - c) = (a + b + c)(2 (ab + bc + ca) - (a^2 + b^2 + c^2))$
$\Leftrightarrow 8 abc - 2 abc + a^2 (b + c) + b^2 (c + a) + c^2 (a + b) - (a^3 + b^3 + c^3)
= 6 abc + a^2 (b + c) + b^2 (c + a) + c^2 (a + b) - (a^3 + b^3 + c^3) $
which is true.
