Few days ago I recalled an interesting inequality I used in my PhD work. It looks like, for any integrable function $f(x) > 0$, $\forall x\in [a,b]$
$$e^{\frac{1}{b-a}\int\limits_{a}^{b}\ln{f(x)}dx}\leq \frac{1}{b-a}\int\limits_{a}^{b}f(x)dx$$
This is, more or less, a generalised form of the inequality of arithmetic and geometric means. E.g. for any $x_i \in[a,b]$, $\forall i=1..n$, $a=x_1$, $b=x_n$
$$\sqrt[n]{\prod\limits_{i=1}^n f(x_i)}\leq \frac{1}{n} \left(\sum\limits_{i=1}^n f(x_i)\right)$$
The natural logarithm is monotonically increasing function:
$$\frac{1}{n}\left(\sum\limits_{i=1}^n \ln{f(x_i)}\right)\leq \ln{\left(\frac{1}{n} \left(\sum\limits_{i=1}^n f(x_i)\right)\right)}$$
Let's consider $h=\frac{b-a}{n}$
$$\frac{1}{b-a}\cdot\frac{b-a}{n}\left(\sum\limits_{i=1}^n \ln{f(x_i)}\right)\leq \ln{\left(\frac{1}{b-a} \left(\sum\limits_{i=1}^n f(x_i)\cdot\frac{b-a}{n}\right)\right)}$$
Or
$$\frac{1}{b-a} \left(\sum\limits_{i=1}^n h\cdot\ln{f(x_i)}\right)\leq \ln{\left(\frac{1}{b-a} \left(\sum\limits_{i=1}^n f(x_i)\cdot h\right)\right)}$$
Now let's consider $\lim\limits_{h\to 0}$ and the fact that the limit keeps the inequality:
$$\lim\limits_{h\to 0} \frac{1}{b-a} \left(\sum\limits_{i=1}^n h\cdot\ln{f(x_i)}\right) = \frac{1}{b-a}\int\limits_{a}^{b}\ln{f(x)} dx$$
The natural logarithm function is continuous, so:
$$\lim\limits_{h\to 0} \ln{\left(\frac{1}{b-a} \left(\sum\limits_{i=1}^n f(x_i)\cdot h\right)\right)}=\\
\ln{\left(\frac{1}{b-a} \cdot \lim\limits_{h\to 0} \left(\sum\limits_{i=1}^n f(x_i)\cdot h\right)\right)}=
\ln{\left(\frac{1}{b-a} \cdot \int\limits_{a}^{b}f(x)dx\right)}$$
As a result
$$\frac{1}{b-a}\int\limits_{a}^{b}\ln{f(x)} dx \leq \ln{\left(\frac{1}{b-a} \cdot \int\limits_{a}^{b}f(x)dx\right)}$$
Considering that the natural exponential function is also monotonically increasing function we receive the original inequality.
Another proof is the fact that the natural logarithm function is concave, i.e. for $\forall \alpha_i$ such that $\sum\limits_{i} \alpha_i=1$
$$\sum\limits_{i} \alpha_i \cdot \ln{f(x_i)} \leq \ln{\left(\sum\limits_{i} \alpha_i \cdot f(x_i)\right)}$$
Now, if we consider $\alpha_i=\frac{Δx_i}{b-a}=\frac{h}{b-a}$ we will receive the same result.
