User:Kompik/ExpIneq


 * $$e^{\frac{x+y}2}\le \frac{e^y-e^x}{y-x} \le \frac{e^x+e^y}2$$

holds for $$x0$$

Using Jensen
The left inequality can be shown using Jensen's inequality.

We have
 * $$\frac{\int_a^b f(x) \,\mathrm{d}x}{b-a} \ge f\left(\frac{a+b}2\right)$$

for any convex function. (This says that for any convex function, the average value is $$\ge$$ than the value in the middle.)

For $$f(x)=e^x$$ we get:
 * $$\frac{e^b-e^a}{b-a} \ge e^{\frac{a+b}2}.$$

Integration
If $$f$$ is a convex function then
 * $$f(ta+(1-t)b) \le t f(a)+ (1-t) f(b)$$
 * $$\int_0^1 f(ta+(1-t)b) \,\mathrm{d}t \le \int_0^1 t f(a)+

(1-t) f(b) \,\mathrm{d}t$$
 * $$\frac{\int_a^b f(x) \,\mathrm{d}x}{b-a} \le f(a) \int_0^1 t

\,\mathrm{d}t + f(b) \int_0^1 (1-t) \,\mathrm{d}t$$
 * $$\frac{\int_a^b f(x) \,\mathrm{d}x}{b-a} \le

\frac{f(a)+f(b)}2$$

Using this for $$f(x)=e^x$$ we get the right inequality.

Note that we have, for any convex function f, the following
 * $$f\left(\frac{a+b}2\right)\le \frac{\int_a^b f(x)

\,\mathrm{d}x}{b-a} \le \frac{f(a)+f(b)}2$$