Relationship between convexity and superadditivity?

Convexity and superadditivity: If a function $f$ is convex, increasing, and $f(0)=0$, then $f$ is superadditive. To see this, draw the secant line through $(0,0)$ and $(x+y,f(x+y))$. This is the graph of a linear (additive!) function $\ell(x)$. By convexity, $f(x)\le \ell(x)$ and $f(y)\le \ell(y)$, hence the superadditivity.

Superadditivity does not imply convexity. To see this, take a convex function such as $f(x)=x^2$ and cut off a small piece of its graph: say, $\tilde f(x)= \max(x^2,2x-1+\epsilon)$ where $\epsilon >0$ is small. This is no longer strictly convex (the graph contains a line segment), but is still strictly superadditive (the line segment is too short to contain $x,y,x+y$). Now replace the line segment with a slightly concave curve, and you have a non-convex superadditive function.

Generalized Hölder's inequality follows by induction from the ordinary one. Let $Y=X_1\cdots X_{n-1}$. Then
$$\mathbb E(YX_n)\le \| Y\|_q\|X_n\|_{p_n}\tag1$$ where $q=p_n/(p_n-1)$. Write $\tilde p_i=p_i/q $ and note that $$\sum_{i=1}^{n-1} 1/\tilde p_i =(1-1/p_n)/q=1 \tag2$$ By the inductive hypothesis, $$ \| Y\|_q^q = \mathbb E(|Y|^q)\le \prod_{i=1}^{n-1}\||X_i|^q\|_{\tilde p_i} =\prod_{i=1}^{n-1}\|X_i \|_{ p_i}^q \tag3 $$ which together with (1) completes the proof.

Jensen's inequality: To me, it is a statement about the relation between convex functions and affine functions: members of the former family are formed by taking supremum of the latter. The proof of inequality expresses this relation better than the inequality itself. Also, this quote from Gowers comes to mind:

One will not get anywhere in graph theory by sitting in an armchair and trying to understand graphs better.