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In a high school reference book, I read a question asking which of $\frac{1+e^x}2$ and $\frac{e^x-1}x$ is larger when $x>0$.

Of course, the question is technically easy to answer. E.g. using the power series expansion of $e^x$, one immediately sees that the average height is greater than the slope of the secant line. Alternatively, one can answer the question by using the first and second derivatives of $x(1+e^x)-2(e^x-1)$.

Yet, answers like these seem too hacky. A more elegant one is to compare the derivatives of $\frac x2$ and $\tanh\frac x2=\frac{e^x-1}{e^x+1}$, but I don't expect a high school student to be aware of the use of hyperbolic tangent.

Since both $\frac{1+e^x}2$ and $\frac{e^x-1}x$ have geometric interpretations, I wonder if there is a more natural --- and probably geometrically minded --- answer. Any idea?

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    $\begingroup$ so $0<x<2$ right ?because $x>2$ seems to be obvious. $\endgroup$ Mar 8, 2018 at 2:19
  • $\begingroup$ @SeyhmusGüngören Actually I didn't notice that. Thanks for the simplification! $\endgroup$
    – user1551
    Mar 8, 2018 at 2:21
  • $\begingroup$ The inequality is true for $x\ne0$, not only $x\gt0$. $\endgroup$
    – robjohn
    Mar 9, 2018 at 9:57

6 Answers 6

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You might be interested in the strict formulation of the Hermite–Hadamard inequality—if $f$ is strictly convex on $[a,b]$ where $a<b$, then the integral of $f$ is strictly bounded above by its trapezoidal approximation: $$\int_a^bf(x)\mathrm{d}x< \tfrac{f(a)+f(b)}{2}(b-a)\text{.}$$ Set $a=0$, $b=x$, $f(x)=\mathrm{e}^x$. Then $$\mathrm{e}^x-1< x\left(\tfrac{\mathrm{e}^x+1}{2}\right)\text{.}$$ Since $f$ in this case is positive and smooth, it's pretty easy to illustrate the inequality graphically.

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  • $\begingroup$ That's very nice indeed. +1 $\endgroup$
    – user1551
    Mar 8, 2018 at 2:30
  • $\begingroup$ user1551 : I just realized that you asked for a strict inequality. Edited accordingly. $\endgroup$
    – K B Dave
    Mar 8, 2018 at 2:47
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$$\frac{1+e^x}2>\frac{e^x-1}x$$ is equivalent to $$x+2>(2-x)e^x $$

These two functions, $x+2$ and $ (2-x)e^x $ have the same value and the same slope at $x=0$

for $x>0$, comparing the two derivatives, $1$ and $(1-x)e^x$ leads to comparing $ \frac {1}{1-x}$ and $e^x$

Note that on the interval $(0,1)$, $\frac {1}{1-x}= 1+x+x^2+...$ wins over $e^x = 1+x+x^2/2+x^3/6+..$

For $x\ge 1$ the $1>(1-x)e^x$ is straight forward.

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Since we are interested in $0<x<2$, make the change $0<x=\ln t<2 \Rightarrow 1<t<e^2$. The inequality will take the form: $$\frac12 \ln t>\frac{t-1}{t+1}$$ At $t=1$ they are equal. The LHS function grow faster: $$\frac{1}{2t}>\frac{2}{(t+1)^2} \iff (t-1)^2>0.$$

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  • $\begingroup$ This change of variables does tidy things up. +1 $\endgroup$
    – user1551
    Mar 8, 2018 at 7:26
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Not a proof but a geometric interpretation: let $$\begin{align}O&=(0,0)&A&=(1,0)&P&=(x,y)\end{align}$$ where $x^2-y^2=1$ is the unit hyperbola and $x>0$. Let $B$ be the intersection of the line $x=1$ with $OP$. Then $$B=(1,\tfrac{y}{x})\text{.}$$ Now, triangle $\triangle OAB$ is contained completely within the hyperbolic sector $OAP$, so we have $$\text{Area}(\triangle OAB)<\text{Area}(OAP)\text{.}$$ Suppose the area of sector of $OAP$ is given by $\tfrac{\tau}{2}$. Then $$(x,y)=(\cosh\tau,\sinh\tau)\text{,}$$ and the area inequality above is $$\tfrac{1}{2}\tanh \tau < \tfrac{\tau}{2}\text{.}$$

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  • $\begingroup$ Nice geometric interpretation. +1. But the fact that the area of the hyperbolic sector is $\tau/2$ may be inobvious. That said, I think it's a good alternative answer to your other one. $\endgroup$
    – user1551
    Mar 8, 2018 at 7:58
  • $\begingroup$ I agree, it's not a good fit for modern definitions. I included it to illustrate that a student who understands $\tfrac{1}{2}\sin \theta\cos\theta<\tfrac{\theta}{2}<\tfrac{1}{2}\tan \theta$ can understand $\tfrac{1}{2}\tanh \tau < \tfrac{\tau}{2}<\tfrac{1}{2}\sinh\tau\cosh\tau$ even if they do not know any calculus. $\endgroup$
    – K B Dave
    Mar 8, 2018 at 8:59
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The inequality is equivalent to $$\tanh{x\over2}={e^x-1\over e^x+1}<{x\over2}\qquad(x>0)\ .$$ As $\tanh0=0$ and $\>\tanh' u=1-\tanh^2 u<1$ $\>(u>0)$ the claim follows.

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If $x\ne0$, then $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\frac{e^x-1}{e^x+1} &=\frac{\mathrm{d}}{\mathrm{d}x}\left(1-\frac2{e^x+1}\right)\\[3pt] &=\frac{2e^x}{\left(e^x+1\right)^2}\\[3pt] &=\frac2{4+\left(e^{x/2}-e^{-x/2}\right)^2}\\ &\in\left(0,\frac12\right) \end{align} $$ Therefore, the Mean Value Theorem says $$ \frac{\frac{e^x-1}{e^x+1}-0}{x-0}\in\left(0,\frac12\right) $$ That is, $$ \frac{\ \frac{e^x-1}x\ }{\ \frac{e^x+1}2\ }\in(0,1) $$

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