pr = (p - a)r_a = (p - b)r_b = (p - c)r_c, where

p = (a + b + c)/2; r is the inradius and r_a, r_b, r_c are exradii of the triangle.

As you may realise the quantities p-a, p-b, p-c are the radii of the Soddy circles.

If you can't get hold of the book (which is a jewel addition to every geometry lover's library) I'll try to find time to put the proof on the web.

Relatively expensive, the book is available from amazon.com:

https://www.amazon.com/Introduction-Geometry-2nd-H-Coxeter/dp/0471504580/sr=1-1/qid=1171807024/ref=pd_bbs_sr_1/102-7772941-9832935?ie=UTF8&s=books

< Has anyone ever thought of the Descartes problem rediscovered by F. Soddy? Given three mutually tangent circles there is a smaller internal fourth circle tangent to all three. I have not found anywhere the proof to the beautiful formula that is just given to all internet sit's as granted, but never proven. I have found a proof on pages 157-158 of Dan Pedoe's Geometry but I admit it is utterly incomprehensible to me (what could the product of two circles mean?)

Do you know of a solution perhaps with the use of trigonometry and plane geometry that I could understand? I presume the Descartes problem is the continuation of the Malfatti triangle, i.e. given three mutually tangent circles calculate the sides of the triangle enclosing all those three circles. >

I am going to give you a visual , maybe sketchy but straightforward answer to your question and a method to draw the two circles tangents to our three given ones.
The method I use is a kind like your ancient Greek ancestors could had done it, no algebra no trigonometry involved just with the help of straightedge and compass

It is based on the inversion, the procedure is at follows.

1) Given the three mutually tangents circle (c0) , (c1) and (c2) draw a line joining the centers of two of them lets say (c1) and (c2).

2) At origin O the tangent point of (c1) and (c2) draw a circle that pass on the center of our third circle (c0). This circle will be our circle of inversion (I)

3) Circle (I) will intersect (c0) in points i1 and i2. Circle (I) will intersect also (c1) and (c2) in point c1a, c1b and c2a, c2b respectively. Now if we draw straight lines (C’1) and (C’2) passing c1a, c1b and , c2a, c2b this two straight lines will represent the inverses of the circles (c1) and (c2), obviously they are parallel since the inverse of the tangent point of (c1) and (c2) O is at infinite.

4) Now we draw a line joining origin O with m the tangent point between circle and (c1), we will extend the line till cut the straight line (C’1) at M’. Obviously M’ will represent the inverse of m. We do the same with circle (c2) and we get point N’ also inverse of tangent point n.

5) Since the intersection points of circle (c0) with circle (I) i1 and i2 remain unchanged in the inversion our circle (c0) will be transformed in circle (C’0) passing through i1, i2 M’ obviously will also pass over N’.

6) Now it is clear that you can draw two circle tangent simultaneously to (C’0) (C’1) and (C’2), one above and one below of (C’0). The two circles called (B’1) and (B’2) are the inverse circles of the sought ones. In the drawing attached is not represented (B’2), not to mess with too many lines.

7) Now to construct (b1) inverse of (B’1) we will draw lines OQ’ and OR’ those two lines will cut respectively (c1) and (c2) in r and q the transformed points of R’ and Q’. We also draw the line OP’ being P’ the tangent point of circles (B’1) and (C’0) this line will render point p the transformed of P’, obviously p must belong to circle (c0).

8) Finally we draw a circle passing for pints p, q and r this circle is the inner solution. For the outer solution one can be obtained by following the same procedure with circle (B’2) not represented.