$\tan^{-1}x$, $\tan^{-1}y$, $\tan^{-1}z$ are in arithmetic progression, as are $x$, $y$, $z$. Show ... 
$\tan^{-1}x, \tan^{-1}y, \tan^{-1}z $ are in arithmetic progression, as are $x$, $y$, $z$. (We assume $y \ne 0,1,-1$.) Show:

*

*$x$, $y$, $z$ are in geometric progression

*$x$, $y$, $z$ are in harmonic progression.

*$x=y=z$

*$(x-y)^2 +(y-z)^2+(z-x)^2 =0$


My attempt:
$$\text{A}=\tan^{-1}x \qquad \text{B}=\tan^{-1}y \qquad \text{C}=\tan^{-1}z$$
$$x=\tan A \qquad y=\tan B \qquad  z=\tan C $$
$$x+z=2y$$
$$A+C=2B$$
$$\tan(A + B + C)=\frac{\tan A +\tan B +\tan C - \tan A\tan B\tan C }{1-\tan A\tan B -\tan B\tan C -\tan C\tan A}$$
$$\tan(3B)=\frac{x +y +z - xyz }{1-xy -yz -zx}$$
How do I continue from here?

(Not an answer so much as an elaborate comment ...) 
For fun, I made a "trigonograph" to see why we might expect the elements to be equal (item (3)). Here, the arithmetic progressions are $\alpha$, $\gamma$, $\beta$ and $\tan\alpha$, $\tan\delta$, $\tan\beta$. (Also, I take all values to be positive, and assume $\alpha + \beta < 180^\circ$.)


$$\begin{align}
2 \gamma &\;=\; \alpha + \beta \\[4pt]
2 \tan\delta &\;=\; \tan\alpha + \tan\beta
\end{align}\qquad\implies\qquad
\begin{array}{c}
\gamma \;\leq\; \delta \\
\text{with equality when and only when} \\
\alpha = \beta \;\left(\; = \gamma = \delta\;\right)
\end{array}$$

