Proving the natural log inequality $\frac{3}{4} < \frac{1}{\sqrt{3}} \log(2 + \sqrt{3}) < \frac{\pi}{4}$ I stumbled upon the following sharp inequality
$$
    \frac{3}{4} < \frac{1}{\sqrt{3}} \log(2 + \sqrt{3}) < \frac{\pi}{4}
$$ 
I got a bit progress on the first part, but the second inequality won't budge.
Any tips on how to prove it would be more than welcome.

A good starting point is:
$$\begin{eqnarray*}\log(2+\sqrt{3})&=&\int_{0}^{\pi/3}\frac{d\theta}{\cos\theta}=2\int_{0}^{1/\sqrt{3}}\frac{dt}{1-t^2}\\&=&\frac{2}{\sqrt{3}}\sum_{j=0}^{+\infty}\frac{1}{(2j+1)\,3^j}\end{eqnarray*}$$
so we just have to prove that:
$$\frac{3}{4}<\sum_{j=0}^{+\infty}\frac{2}{(2j+1)\,3^{j+1}}<\frac{\pi}{4}.\tag{1}$$
The LHS inequality is trivial (it follows by summing the terms corresponding to $j=0,1,2$), while the RHS follows from the fact that the series converges pretty fast:
$$\sum_{j=0}^{+\infty}\frac{8}{(2j+1)3^{j+1}}\leq\sum_{j=0}^{2}\frac{8}{(2j+1)\,3^{j+1}}+\sum_{j=3}^{+\infty}\frac{8}{7\cdot 3^{j+1}}=3+\frac{1}{45}+\frac{4}{189}<3+\frac{4}{90}<\pi.$$
