Intuitive proof of $\frac{1+e^x}2>\frac{e^x-1}x$ for high school students

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.

$$\frac{1+e^x}2>\frac{e^x-1}x$$ 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.

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: $$\frac{1}{2t}>\frac{2}{(t+1)^2} \iff (t-1)^2>0.$$