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

 A: To prove that $$\tan(x+\frac{\pi}8)>e^x+\ln x \qquad\ x\in(0,\frac{3\pi}8)$$ I suppose that it is sufficient to show that function $$F(x)=\tan(x+\frac{\pi}8)-e^x-\ln x $$ is always positive in the given range.
The function is positive infinite at both ends. So, let us show that $F(x)$ goes through aminimum value which is positive. Compute the derivatives $$F'(x)=\sec ^2\left(x+\frac{\pi }{8}\right)-e^x-\frac{1}{x}$$ $$F''(x)=2 \tan \left(x+\frac{\pi }{8}\right) \sec ^2\left(x+\frac{\pi
   }{8}\right)-e^x+\frac{1}{x^2}$$ The first derivative is $-\infty$ at the left bound and $\infty$ at the upper bound. If you plot $F'(x)$, you will notice that $F'(x)=0$ has a single root.
Since no explicit solution can be obtained for such as a transcendental equation, we need to use some numerical method. So, let us use Newton method will will give for the iterates $$x_{n+1}=x_n-\frac{F'(x_n)}{F''(x_n)}$$ and start iterating at the midpoint of the interval $(x_0=\frac{3 \pi }{16})$. This will give the following iterates
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 0.5890486225 \\
 1 & 0.6131825159 \\
 2 & 0.6121561606 \\
 3 & 0.6121539948 
 \end{array}
\right)$$
Now, using your pocket calculator, $$F(0.6121539948)\approx  0.22053$$ $$F''(0.6121539948)\approx 11.7739$$ The second derivative test confirms that the point is a minimum.
A: 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
A: 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.
