Q:

If the sides $a, b, c $ form a successive geometric progression with common ratio $r(r>1)$, then prove that $A<B<\frac{\pi}{3}<C$

From the given data, the obvious result that can be drawn is 
$$b^2=ac$$
Also $$\cos B =\frac{a^2+c^2-b^2}{2ac}$$
$$\cos B =\frac{a^2+c^2}{2b^2}-\frac{b^2}{2b^2}$$
$$\cos B =\frac{(a+c)^2}{2b^2}-1-\frac 12$$
I don’t know how to proceed.

A:

You are so closeee
$$
\begin{aligned}
\frac{a^{2}+c^{2}-b^{2}}{2ac}&=\frac{a^{2}+c^{2}-ac}{2ac}\\
&>\frac{2ac-ac}{2ac}
\end{aligned}
$$
Thus $\angle A<\angle B < \frac{\pi}{3}$

