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Show that for $a \neq b$ it holds: $$\frac{e^b-e^a}{b-a} < \frac{e^b+e^a}{2}$$

My first idea was to rearrange $$2 \cdot (e^b-e^a) < (b-a)(e^b+e^a)$$

$$2e^b-2e^a < be^b + be^a - ae^b -e^a$$

And then take the natural logarithm on both sides $$\ln{(2e^b-2e^a)} < \ln{(be^b + be^a - ae^b -e^a)}$$

I'm a bit lost at this point

PapuaNewGuinea
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  • Assume that $b>a$. You could then start with $2\mathrm{e}(\mathrm{e}^{b-a}-1) < (b-a)\mathrm{e}(\mathrm{e}^{b-a}+1)$. Substitute $x := b-a.$ Then we have $2(\mathrm{e}^x - 1) < x(\mathrm{e}^x+1).$ At least for $x>2$ this is easy to show. Maybe there is an easy way for the values that are left. – Lereu Nov 25 '23 at 22:16
  • Just thought I'd mention that the inequality is equivalent to ${e^{b-a \over 2} - e^{-{b - a \over 2}} \over e^{b-a \over 2} + e^{-{b - a \over 2} } }< {b - a \over 2}$, so for $y = {b - a \over 2}$ you need $\tanh |y |< |y|$, which follows from the fact that $\tanh 0 = 0$ and has derivative ${\rm sech}^2 y$, which is greater than $1$ for $y \neq 0$. – Zarrax Nov 25 '23 at 22:17

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