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...

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,
$$\begin{align}2\sin^2\frac{\alpha}2&=\frac{4yz}{2(x+z)(x+y)}\\ \sin\dfrac{\alpha}2&=\dfrac1{\sqrt{\dfrac{(x+z)(x+y)}{yz}}}\\&=\frac1x\frac1{\sqrt{\dfrac{(x+z)}{xz}\dfrac{(x+y)}{xy}}}\\&=\frac1x\frac1{\sqrt{\left(\dfrac1x+\dfrac1z\right)\left(\dfrac1x+\dfrac1y\right)}}\end{align}$$
$\square$
