Intuitive proof of $\frac{1+e^x}2>\frac{e^x-1}x$ for high school students 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?
 A: $$\frac{1+e^x}2>\frac{e^x-1}x$$ 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. 
A: 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:
$$\frac{1}{2t}>\frac{2}{(t+1)^2} \iff (t-1)^2>0.$$
A: 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.
A: 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
$$\tfrac{1}{2}\tanh \tau < \tfrac{\tau}{2}\text{.}$$
A: 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,
$$
\frac{\ \frac{e^x-1}x\ }{\ \frac{e^x+1}2\ }\in(0,1)
$$
A: 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.
