I am trying to prove a version of Descartes' theorem in an elementary way.
Given three mutually tangent circles (no one is inside the others) with centres $A,B,C$ and radii $a, b, c$ respectively, I would like to compute the radius of the inner circle which is tangent to all the given circles without using Descartes' theorem. My attempt is as follows.
Let $I$ and $x$ be the centre and radius of the fourth circle. Applying the Heron's formula to triangles $IBC,ICA,IAB$ and $ABC$ and using the fact that the area of the fourth triangles is equal to the sum of the areas of the other three, we get $$\sqrt{bcx(b+c+x)}+\sqrt{cax(c+a+x)}+\sqrt{abx(a+b+x)}=\sqrt{abc(a+b+c)}.$$
Even though the equation above has only one root (the left side is increasing in $x$), it is still difficult to show that the Soddy circles formula $$x=\frac{abc}{bc+ca+ab+2\sqrt{abc(a+b+c)}}$$ gives the unique root. (I tried to substitute that value of $x$ into the previous equation, however it is a mess!)
Can someone help me? Thanks a lot!