Prove that $\tan(x+\frac{\pi}8)>\mathrm{e}^x+\ln x$ for $x\in(0,\frac{3\pi}8)$ 
Prove that $$\tan\left(x+\frac{\pi}8\right)> \mathrm{e}^x+\ln x,\quad x\in\left(0,\frac{3\pi}8\right).$$

By plotting the graph I found that this is indeed true (in fact I found this inequality through plotting), but how can I prove this? There are various functions in this inequality and I don't know how to start. Any hints will be appreciated.
Edit: I was suggested to post the graph (from WolframAlpha). This is the graph of $\tan(x+\frac{\pi}8)-\mathrm{e}^x-\ln x$.


Alternative proof by Tangent Line Method (cf. Lozenges's proof):
Let $f(x)  = \tan(x + \pi/8)$.
Since $f(x)$ is convex on $(0, 3\pi/8)$, we have
$$ f(x)
\ge f(5\pi/24) + f'(5\pi/24)(x - 5\pi/24) = 4x + \sqrt{3} - 5\pi/6.$$
It suffices to prove that, for all $x\in (0, 3\pi/8)$,
$$4x + \sqrt{3} - 5\pi/6 > \mathrm{e}^x + \ln x.$$
It is easy to prove that, for all
$x\in (0, 3\pi/8)$,
$$\mathrm{e}^x \le 1 + x + x^2.$$
Thus, it suffices to prove that
, for all $x\in (0, 3\pi/8)$,
$$4x + \sqrt{3} - 5\pi/6 > (1 +x + x^2) + \ln x$$
which is true (easy to prove).
We are done.
