Convex function strictly increasing I want to prove that if function $f:(a,b)\to \mathbb{R}$ is not constant, increasing and convex then from some $c\in (a,b)$ function $f:[c,b)\to \mathbb{R}$ is strictly increasing. Any ideas?
 A: Lemma. For $\forall \,u \lt v \lt w \in (a,b)\,$:
$$
\frac{f(v)-f(u)}{v-u} \le \cfrac{f(w)-f(v)}{w-v} 
$$
This simply formalizes the intuition that adjacent chords on the graph of a convex function have non-decreasing slopes, and follows directly from the definition of convexity with $\lambda = \frac{w-v}{w-u} \in [0,1]$ and $1-\lambda=\frac{v-u}{w-u}\,$:
$$
\require{cancel}
\begin{align}
& f\big(\lambda u + (1-\lambda)w)\big) \le \lambda f(u) + (1-\lambda)f(w) \\
\iff \quad & f\left(\frac{(\cancel{w}-v)u+(v-\cancel{u})w}{w-u}\right) \le \frac{w-v}{w-u} f(u) + \frac{v-u}{w-u} f(w) \\
\iff \quad & f(v) \big((w-v)+(v-u)\big) \le (w-v)f(u) + (v-u)f(w) \\
\iff \quad & \big(f(v)-f(u)\big)(w-v) \le \big(f(w)-f(v)\big)(v-u)
\end{align}
$$
Proof. Since $f$ is increasing and not constant there exists $\exists \,c \in (a,b)$ such that $f(c) \gt f(a)$. For $\forall \,x_1,x_2$ such that $a\lt c\lt x_1\lt x_2\lt b$ applying the previous lemma twice gives:
$$
\frac{f(x_2)-f(x_1)}{x_2-x_1} \ge \frac{f(x_1)-f(c)}{x_1-c} \ge \frac{f(c)-f(a)}{c-a} \gt 0 
$$
The above proves that $f$ is strictly increasing on $[c,b)$ since $x_1 \gt c \implies f(x_1) \gt f(c)$ and $x_2 \gt x_1 \gt c \implies f(x_2) \gt f(x_1)$.
A: Here is an outline. If the function is not constant, there is some value $c \in [a, b]$ for which $f(c) \ne f(a)$. Since $f$ is increasing and $c > a$, $f(c) > f(a)$. By convexity the graph of $f$ is above the line through $(a, f(a))$ and $(c, f(c))$. Therefore on $[c, b]$, $f$ is strictly increasing.
