Given strongly convex $f(x)$ definded on $[a, b]$. I'm trying to prove that $\forall x \in (a, b)$ correct this strong inequality: $$f(x) < \max \{f(a), f(b) \}. $$ I can understand this fact on intuition level - from geometric side of convexity - every point under the chord, connecting any two points in interval, in which function defined. So each point is under the chord and under the $\max \{ f(a), f(b) \}$.
But how can i formalize it?
UPD: there was an misspell: i need it for strict, not strong convexity.
Recall that a function is strongly convex if and only if there is $m>0$ such that $f(y)\geq f(x)+\nabla f(x)(y-x)+\frac{m}{2}\lVert x-y \rVert^2$ for all $x,y$. Let $z= tx+(1-t)y$ for $0<t<1$, then: $$f(x)-f(z) \geq \nabla f(z)(x-z)+\frac{m}{2}\lVert x-z \rVert^2$$ $$f(y)-f(z) \geq \nabla f(z)(y-z)+\frac{m}{2}\lVert y-z \rVert^2$$ It follows that: $$\max\{f(x), f(y)\} \geq tf(x)+(1-t)f(y) \geq f(z)+ \frac{m}{2}\lVert x-z \rVert^2 +\frac{m}{2}\lVert y-z \rVert^2$$ Hence $f(tx+(1-t)y) < tf(x)+(1-t)f(y)\leq \max\{f(x), f(y)\}$. Note that we also showed that $f$ is strictly convex.