 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. ThusFor strict Jensen's inequality, put $m=(b-a{)}^{-1}{\int }_a^{b}u$. 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 strict integral Jensen inequality.