$$\tan\left(x+\frac{\pi}8\right)>e^x+\ln x,\ x\in\left(0,\frac{3\pi}8\right)$$

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


  • 2
    $\begingroup$ The left-hand side is undefined for $x=\frac38\pi$, and the right-hand side is undefined for $x=0$. Should the range for $x$ have been an open interval instead? $\endgroup$ – Henning Makholm Sep 25 '16 at 4:35
  • $\begingroup$ @HenningMakholm Oh yes, my mistake, sorry. I'll fix it right away $\endgroup$ – Colescu Sep 25 '16 at 4:37
  • $\begingroup$ post the graph? likely just take a derivative and confirm the function is increasing or decreasing. $\endgroup$ – djechlin Sep 25 '16 at 4:47
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    $\begingroup$ @djechlin Have you taken the derivative to discern anything useful here? The function $\tan(x+\pi/8)-e^x-\log(x)$ is neither everywhere increasing nor everywhere decreasing. $\endgroup$ – Mark Viola Sep 25 '16 at 5:00
  • $\begingroup$ Well I need to clarify that this is not a calculus problem. I don't know what methods or techniques are to be used, but I think calculus might help so I hashtagged calculus. $\endgroup$ – Colescu Sep 25 '16 at 5:24

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.


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


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