$a+b+c=-3\sqrt{ac}$ 
If $a,b,c$ be in Geometic Progression, and $b-c,c-a,a-b$ in Harmonic Progression then prove that $a+b+c=-3\sqrt{ac}$.

My attempt: $$c-a=\frac{2(b-c)(a-b)}{b-c+a-b}$$$$(c-a)^2+2(b-c)(a-b)=0$$$$c^2+a^2-2ac+2(ab-b^2-ac+bc)=0$$$$a^2+c^2-2b^2-4ac+2ab+2bc=0$$$$a^2+c^2-6ac+2b(a+c)=0(b^2=ac)$$$$a^2+c^2+2ac+2b(a+c)=8ac$$$$(a+c)(a+c+2b)=8ac$$$$(a+c)(\sqrt{a}+\sqrt{c})^2=8(\sqrt{ac})^2(b=\sqrt{ac})$$

Let $$a,b,c \equiv k,kr,kr^2 \text{ be in GP }  $$
Then $$c-a = \frac{2(b-c)(a-b)}{b-c+a-b}  \implies- (kr^2-k)^2 = 2(kr-kr^2)(k-kr) $$
$$\implies-(r^2-1)^2 = 2(r)(1-r)(1-r) \implies-(r-1)^2(r+1)^2 = 2r(r-1)^2$$
$\color{red}{r\neq1} $ as we have a GP

$$ -(r+1)^2 = 2r \text{ or } r^2+1 =- 4r$$

Now $$a+b+c = k(1+r+r^2) = k(-4r+r) = -3kr$$ and $-3\sqrt{ac} = -3\sqrt{k^2r^2} = -3|kr| =\begin{cases}  -3kr \iff kr>0 \\ 3kr \iff kr<0 \end{cases}$ 
So if $kr>0$ we have

$$a+b+c = -3\sqrt{ac} = -3kr$$

Edit : There's a possibility that $k$ and $r$ can both be negative. So, I've used $kr>0$ 
