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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!!

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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.

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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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