Prove that $r^2+ r_1^2+ r_2^2 +r_3^2=16R^2-a^2-b^2-c^2$

Question

In any $∆ABC$, prove that: $$r^2+ r_1^2+ r_2^2 +r_3^2=16R^2-a^2-b^2-c^2$$

where $r$ is the inradius, $r_1,r_2,r_3$ are the exradii of the triangle respectively, and $R$ is the circumradius of the triangle.

I tried to solve the question by replacing $r$ by $∆/s$ and $r_1$ by $∆/(s-a)$, and similarly for $r_2$ and $r_3$, but the answer is too long and I think there must be another short method.

Answer 1

Hint: Use $\displaystyle(r_1+r_2+r_3-r)^2=r_1^2+r_2^2+r_3^2-2r(r_1+r_2+r_3) +2(r_1r_2+r_2r_3+r_3r_1)$ knowing that $$r_1+r_2+r_3-r=4R, \\ r_1r_2+r_2r_3+r_3r_1=s^2$$