= Solution
A <strictly convex function> satisfies the same inequality strictly whenever $x\ne y$ and $0<t<1$.
For the unlabelled Hessian criterion, <convexity> implies $F(z_0+th)+F(z_0-th)-2F(z_0)\geq0$. Divide by $t^2$ and let $t\to0$ to obtain $h^TD^2F(z_0)h\geq0$ for every $h$. Thus the <Hessian matrix> is <positive semidefinite>.
Conversely, if every Hessian is <positive semidefinite>, Taylor's <integral> formula with $h=z-z_0$ gives
$$
F(z)-F(z_0)-DF(z_0)\cdot h=\int_0^1(1-t)h^TD^2F(z_0+th)h\,dt\geq0.
$$
The allowed supporting-plane characterization then proves
$$
\boxed{F\text{ convex}\iff D^2F(z)\text{ positive semidefinite for all }z.}
$$
For the <integral> inequality put $m=(b-a)^{-1}\int_a^bu$. The supporting tangent at $m$ gives $F(u(x))\geq F(m)+F'(m)(u(x)-m)$ <almost everywhere>. Integrating cancels the linear term, so
$$
\boxed{F\left(\frac1{b-a}\int_a^bu\right)\leq\frac1{b-a}\int_a^bF(u).}
$$
This is <Jensen's inequality>. It remains valid with value $+\infty$ on the right: the affine lower bound has <integrable> negative part, so the extended <integral> is well-defined.
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