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 $\tan(x+\frac{\pi}8)-\mathrm{e}^x-\ln x$.


We show that there exists $m\in \mathbb{R}^+$ such that for all $x\in \left(0,\frac{3\pi }{8}\right)$ 
$$\tan \left(x+\frac{\pi }{8}\right)>m\left(x-\frac{\pi }{16}\right)>e^x+\ln  x$$
$m$ is the slope of the tangent line to 
$$f(x)=\tan \left(x+\frac{\pi }{8}\right)$$
at $x=a$ and passing through the point $\left(\frac{\pi }{16},0\right)$
Since $f(x)$ is concave up in the interval we have
$$f(x)>m\left(x-\frac{\pi }{16}\right)$$
for $x\in \left(0,\frac{3\pi }{8}\right)$
Now, $a$ satisfies the equation
$$\sec ^2\left(a+\frac{\pi }{8}\right)=\frac{\tan \left(a+\frac{\pi }{8}\right)}{a-\frac{\pi }{16}}$$
which simplified gives
$$2a-\frac{\pi }{8}=\sin \left(2a+\frac{\pi }{4}\right)$$
From this we get $a\approx 0.637575$ and
$$m=\sec ^2\left(a+\frac{\pi }{8}\right)\approx 3.77648$$
Since $g(x)=e^x+\ln  x$ is strictly increasing over the interval, to show that 
$$g(x)<m\left(x-\frac{\pi }{16}\right)$$
it suffices to verify the inequality at $x=\frac{3\pi }{8}$
Check
$$\frac{g\left(\frac{3\pi }{8}\right)}{\frac{5\pi }{16}}\approx 3.47552 <3.77648\approx m$$
Therefore, $g(x)<f(x)$ over the given interval
