Distance between the point of touching in three touching circles $\textbf{Question : }$ Say you have three touching circles $\Gamma_1,\Gamma_2,\Gamma_3$ with radii $x,y,z$ and centers $A,B,C$ as per the diagram, then prove the following $$|DE|=\frac{2}{\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x}+\dfrac{1}{z}\right)}}$$

$\textbf{My Attempt}$
Let $\angle{DAE} = \alpha, \angle{FBE}=\beta , \angle{FCD}=\pi-\alpha-\beta$ then through sine rule in the $\Delta ABC$ we can say
$$\frac{\sin{\alpha}}{y+z}=\frac{\sin{\beta}}{z+x}=\frac{\sin{(\alpha+\beta)}}{x+y} $$
As we know
$$|DE|=2x\sin\frac{\alpha}{2}$$
We just need to prove
$$\sin\frac{\alpha}{2} = \dfrac{1}{x}\dfrac{1}{\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x}+\dfrac{1}{z}\right)}}$$
So using the sine rule relation
$$\sin\beta = \frac{z+x}{y+z} \sin \alpha$$
And again from the sine rule relation
$$\sin \alpha = \frac{y+z}{x+y} \sin (\alpha+\beta)$$
Now substituting $\sin \beta$ in terms of $\sin \alpha$
$$\sin \alpha = \sin \alpha \cdot \sqrt{1-\left( \frac{z+x}{y+z}\sin \alpha \right)^2}+\sqrt{1-\sin^2 \alpha}\cdot \frac{z+x}{y+z}\sin \alpha  $$
$$1 =  \sqrt{1-\left( \frac{z+x}{y+z}\sin \alpha \right)^2}+\sqrt{1-\sin^2 \alpha}\cdot \frac{z+x}{y+z}$$
Now this is the step where I get stuck, it is simply too complicated to solve by hand for me at least. Maybe someone can suggest a way to solve this in reasonable time or even better a different approach...
 A: From double angle formulae, we have $$2\sin^2\frac{\alpha}2=1-\cos\alpha.$$
Using cosine rule,
$$\cdots=1-\frac{b^2+c^2-a^2}{2bc}=\frac{a^2-(b-c)^2}{2bc}=\frac{(a-b+c)(a+b-c)}{2bc}$$
where $a=y+z,\ b=x+z,\ c=x+y$ in your diagram.
Thus,
$$\begin{align}2\sin^2\frac{\alpha}2&=\frac{4yz}{2(x+z)(x+y)}\\ \sin\dfrac{\alpha}2&=\dfrac1{\sqrt{\dfrac{(x+z)(x+y)}{yz}}}\\&=\frac1x\frac1{\sqrt{\dfrac{(x+z)}{xz}\dfrac{(x+y)}{xy}}}\\&=\frac1x\frac1{\sqrt{\left(\dfrac1x+\dfrac1z\right)\left(\dfrac1x+\dfrac1y\right)}}\end{align}$$
$\square$
A: Partial solution
Set $\gamma=\frac{z+x}{y+z}$ and $\mu=\sin^2\alpha$. You end up with $$1=\sqrt{1-\gamma^2\mu}+\sqrt{\gamma^2-\gamma^2\mu}$$
Squaring gives $$1=(1+\gamma^2)-2\gamma^2\mu+2\sqrt{(1-\gamma^2\mu)(\gamma^2-\gamma^2\mu)}$$
Which can be re-written to $$-\gamma^2+2\gamma^2\mu=2\sqrt{\gamma^2-(1+\gamma^2)\mu+\gamma^4\mu^2}$$
Squaring again gives $$\gamma^4-4\gamma^4\mu+4\gamma^4\mu^2=4\gamma^2-4(1+\gamma^2)\mu+4\gamma^4\mu^2$$
Which simplifies to a linear equation in $\mu$, hurray! We get:
$$\mu=\frac{4(\gamma^2-\gamma^4)}{4(1+\gamma^2-\gamma^4)}=\frac{1-\gamma^2}{\gamma^{-2}+1-\gamma^2}$$
Now we only need to solve $\sin\frac\alpha2$ from $\mu=\sin^2\alpha$.
Set $\sin^2\frac\alpha2=\nu$. We get
$$\mu=\sin^2\alpha=4\sin^2\frac\alpha2(1-\sin^2\frac\alpha2)=4\nu-(2\nu)^2=-((2\nu-1)^2-1)=1-(2\nu-1)^2,$$ so $\nu=\frac{1\pm\sqrt{1-\mu}}2$ and $$\sin\frac\alpha2=\sqrt{\frac{1-\sqrt{1-\mu}}2}=\sqrt{\frac{1-\sqrt{1-\frac{1-\gamma^2}{\gamma^{-2}+1-\gamma^2}}}2} = \sqrt{\frac{1-\frac{\gamma^{-1}}{\sqrt{\gamma^{-2}+1-\gamma^2}}}2}$$
Can you take over from here?
