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.
