I'm trying to show that if a function is strictly convex
$$f(y + \lambda(y-x))< f(y) + \lambda(f(y)-f(x)) $$
then
$$f(y) > f(x) + \nabla f(x)'(y-x) $$
without appealing to the Hessian.
With regular convexity I was able to do this by passing to a limit, where you get a situation like
$$f(x) - f(y) \ge \frac{f(y + \lambda(x-y)) - f(y)}{\lambda}$$
but if I replace the $\ge $ with $>$ the strict inequality might not survive.
Suppose $f(y) = f(x) + \langle \nabla f(x), (y-x) \rangle$ for some $x \neq y$.
Let $\phi(t) = f(x+t(y-x))-f(x)-t \langle \nabla f(x), (y-x) \rangle$. The above can be written as $\phi(1) = \phi(0)$ and we note that $\phi'(0) = 0$, so we have $\phi(t) = \phi(0)$ for all $t \in [0,1]$, which contradicts $\phi$ being strictly convex.
Hence $f(y) > f(x) + \langle \nabla f(x), (y-x) \rangle$ for all $x \neq y$.