Nope, my bad. This is really s²/(rabc).

The challenge is to prove that

1/aR_A + 1/bR_B + 1/cR_C = s²/rabc.

From

https://www.cut-the-knot.org/Curriculum/Geometry/MixtilinearConstruction.shtml

we know that, say,

1 / R_A = cos²(A/2) / r

so that

1 / aR_A = cos²(A/2) / ar.

Now, cos²(A/2) = (1 + cos A) / 2, thus

1/aR_A + 1/bR_B + 1/cR_C =

1/2r · (1/a + 1/b + 1/c + cos(A)/a + cos(B)/b + cos(C)/c).

By the Law of Cosines,

bc·cos(A) = (b² + c² - a²)/2, etc. From which,

cos(A)/a + cos(B)/b + cos(C)/c = (a² + b² + c²)/2abc.

Further,

1/a + 1/b + 1/c + (a² + b² + c²)/2abc = (a + b + c)² / 2abc.

Combining all together,

1/aR_A + 1/bR_B + 1/cR_C =

1/2r · (a + b + c)² / 2abc =

s² / rabc.