[proofplan] We use the supporting line characterisation of convex functions. At the point $m = \mathbb{E}[X]$, there exists an affine minorant $a + bt \leq \phi(t)$ with equality at $t = m$. Applying this inequality pointwise to $X(\omega)$ and taking expectations gives the result. [/proofplan] [step:Construct the supporting line at $m = \mathbb{E}[X]$] Since $\phi$ is convex on $I$ and $m = \mathbb{E}[X] \in I$ (as $I$ is an interval and $X$ takes values in $I$), there exist constants $a, b \in \mathbb{R}$ such that \begin{align*} \phi(t) \geq a + bt \quad \text{for all } t \in I, \end{align*} with equality at $t = m$, i.e., $\phi(m) = a + bm$. This is the supporting hyperplane theorem for convex functions; for interior points of $I$ it follows from the existence of one-sided derivatives. [/step] [step:Apply the supporting line inequality pointwise to $X(\omega)$] For every $\omega \in \Omega$, $X(\omega) \in I$, so \begin{align*} \phi(X(\omega)) \geq a + bX(\omega). \end{align*} [/step] [step:Take expectations and use linearity to conclude] Since $\phi \circ X$ is integrable by hypothesis, taking expectations of both sides (using linearity) gives \begin{align*} \mathbb{E}[\phi(X)] \geq a + b\mathbb{E}[X] = a + bm = \phi(m) = \phi(\mathbb{E}[X]). \end{align*} [/step]