Prove that $r_1^2+r_2^2+r_3^2+r^2=16R^2-a^2-b^2-c^2$
Where r denotes the respective in-radius and R the circum radius
a,b,c are the sides of the triangle
I have tried using the
$\frac{\Delta}{(s-a)\dots}$
But I get a fraction with a irreducible $\frac{denominator}{numerator}$.
$$(r_1 + r_2 + r_3 - r)^2={r_1}^2 + {r_2}^2 + {r_3}^2 +{ r}^2 -2r(r_1 + r_2 + r_3 ) +2(r_1r_2 +r_2r_3+r_3r_1)$$ $$r_1 + r_2 + r_3 - r =4R$$ $$r_1r_2 +r_2r_3+r_3r_1=s^2$$ $$\text{Now,}\qquad 2(rr_1 + rr_2 + rr_3 )$$ $$=2\left[\frac{\Delta^2}{s(s-a)}+\frac{\Delta^2}{s(s-b)}+\frac{\Delta^2}{s(s-c)} \right]$$ $$=2\Delta^2\left[\frac{(s-b)(s-c)+(s-c)(s-a)+(s-a)(s-b)}{\Delta^2}\right]$$