Supposedly the Hermite-Hadamard inequality is quite classical, as its name would suggest, but I hadn’t encountered it until browsing a probability book the other day. Well, what what does this inequality say? If is a convex function, then

The reader may ask how we know that is integrable (in the Riemann or Lebesgue sense). Recall from my previous post on convexity that a convex function defined on an interval of is continuous, hence bounded. So integrability is not a problem.

We now turn to the proof of this inequality. First, the RHS. For , we can write

so by convexity,

Integrating both sides over , we obtain the inequality

Recall that a convex function has left- and right-derivatives at each point of its domain:

Using the nondecreasing monotonicity of the function in both arguments, we obtain

for all and

for all $a \leq x < \frac{a+b}{2}$. Integrating the first inequality over and the second inequality over , we see that

and

Summing these two integrals, we obtain the inequality

since .

We now consider some examples of applications of the Hermite-Hadamard Inequality to calculus. The first example concerns the proof of Stirling’s formula:

Consider the real-valued function . is convex since . Applying the Hermite-Hadamard inequality with , we obtain

The preceding inequalities can be used to establish for a positive integer that

In our second example, let . It is evident that is strictly convex. Hence, for ,

Taking , for positive , we see that

which yields the geometric, logarithmic, and arithmetic mean inequalities. For more on the Hermite-Hadamard inequality, including extensions and applications, see the paper “Old and New on the Hermite-Hadamard Inequality” by Niculescu and Persson.

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