= Solution
A <strictly convex function> satisfies that inequality strictly whenever $x\ne y$ and $0<t<1$.
For necessity of the strict supporting-tangent property, <convexity> gives a nonnegative gap along a segment from $z_0$ to $z$. If the gap at its other endpoint were zero, <convexity> of the gap and its zero initial value would force it to be zero all along the segment. That would make $F$ affine on this nontrivial segment, contrary to <strict convexity>. Conversely let $z=(1-t)x+ty$ with distinct endpoints. Apply the strict supporting inequality at $z$ to both endpoints, multiply by $1-t$ and $t$, and add. The linear terms cancel, yielding <strict convexity>. Thus
$$
\boxed{F\text{ strictly convex}\iff F(z)>F(z_0)+DF(z_0)\cdot(z-z_0)\text{ for }z\ne z_0.}
$$
For strict <Jensen's inequality>, put $m=(b-a)^{-1}\int_a^bu$. The supporting gap $F(u)-F(m)-DF(m)\cdot(u-m)$ is nonnegative everywhere and strictly positive where $u\ne m$. That set has positive measure because $u$ is not <almost everywhere> constant. Its integral is therefore strictly positive. The linear term integrates to zero, giving the requested \b[strict integral <Jensen inequality>].
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