algebra precalculus - In $Delta ABC$ if $a,b,c$ are in Harmonic Progression

In $\Delta ABC$ if $a,b,c$ are in Harmonic Progression Then Prove that

$$\sin ^2\left(\frac{A}{2}\right),\sin ^2\left(\frac{B}{2}\right),\sin ^2\left(\frac{C}{2}\right)$$ are in Harmonic Progression

My Try:

we have

$$\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$$ Then

$$\frac{a-b}{a}=\frac{b-c}{c}$$ and by Sine Rule

$$ \frac{\sin A-\sin B}{\sin A}=\frac {\sin B-\sin C}{\sin C}$$ $\implies$

$$\frac{2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)}{2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)}=\frac{2\sin \left(\frac{A}{2}\right)\cos \left(\frac{B-C}{2}\right)}{2\sin \left(\frac{C}{2}\right)\cos \left(\frac{C}{2}\right)}$$ $\implies$

$$\sin ^2\left(\frac{C}{2}\right)\left(2\cos \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)\right)=\sin ^2\left(\frac{A}{2}\right)\left(2\cos \left(\frac{A}{2}\right)\cos \left(\frac{B-C}{2}\right)\right)$$ $\implies$

$$\sin ^2\left(\frac{C}{2}\right) \left(\sin B+\sin A\right)=\sin ^2\left(\frac{A}{2}\right) \left(\sin B+\sin C\right)$$

Any way to proceed?