Jensen's inequality for integrals What nice ways do you know in order to prove Jensen's inequality for integrals? I'm looking for some various approaching ways.
Supposing that $\varphi$ is a convex function on the real line and $g$ is an integrable real-valued function we have that:
$$\varphi\left(\int_a^b f\right) \leqslant \int_a^b \varphi(f).$$

I like this, maybe it is what you want ...  
Let $E$ be a separable Banach space, let $\mu$ be a probability measure defined on $E$, let $f : E \to \mathbb R$ be convex and (lower semi-)continuous.  Then
$$
f\left(\int_E x d\mu(x)\right) \le \int_E f(x)\,d\mu(x) .
$$
Of course we assume $\int_E x d\mu(x)$ exists, say for example $\mu$ has bounded support.
For the proof, use Hahn-Banach.  Write $y = \int_E x d\mu(x)$.  The super-graph $S=\{(x,t) : t \ge f(x)\}$ is closed convex.  (Closed, because $f$ is lower semicontinuous; convex, because $f$ is convex.)  So for any $\epsilon > 0$ by Hahn-Banach I can separate $(y,f(y)-\epsilon)$ from $S$.  That is, there is a continuous linear functional $\phi$ on $E$ and a scalar $s$ so that $t \ge \phi(x)+s$ for all $(x,t) \in S$ and $\phi(y)+s > f(y)-\epsilon$.  So:
$$
f(y) -\epsilon < \phi(y)+s = \phi\left(\int_E x d\mu(x)\right)+s =
\int_E (\phi(x)+s) d\mu(x) < \int_E f(x) d\mu(x) .
$$
This is true for all $\epsilon > 0$, so we have the conclusion.

