As all you masters in math know, the famous AM-GM inequality in positive numbers is as follows:
$${1\over n}\sum_{k=1}^{n}x_k\ge \left(\prod_{k=1}^{n}x_k\right)^{1\over n}$$
which can be rewritten in the following alternative form:
$$\large{1\over n}\sum_{k=1}^{n}x_k\ge e^{{1\over n}\sum_{k=1}^{n}\ln x_k}$$
An interpretation says that the (ordinary or arithmetic) average of a bunch of data, $x_k$, is not smaller than the exponential of the average of $\ln x_k$s.
Now, if we interpret $x_k$s as infinitesimal parts of some function ,say $f(x)$, and correspondingly consider $\ln x_k$ as infinitesimal parts of $\ln f(x)$, intuition says:
$$\large{1\over T}\int_0^T f(x)dx\ge e^{{1\over T}\int_0^T \ln f(x)dx}$$in an arbitrary interval $(0,T)$.
so naturally, my question is: is this relation even true for any positive $f(x)$? If so, what is a formal proof for that?
The relation is true, it is Jensen's inequality $$ \varphi \left({1\over T}\int_0^T f(x) \, dx \right)\le {1\over T}\int_0^T \varphi( f(x)) \, dx \, , $$
applied to the convex function $\varphi(t) = - \ln t$.