A function $f: \mathbb{R} \to \mathbb{R}$ is convex if$$f(\lambda x + (1 - \lambda)y) \le \lambda f(x) + (1 - \lambda)f(y)$$whenever $x < y \in \mathbb{R}$ and $\lambda \in [0, 1]$.
If $f$ is convex and $x \in \mathbb{R}$, does there exist a real number $c$ such that $f(y) \ge f(x) + c(y - x)$ for all $y \in \mathbb{R}$?
the inequality is true even if $f$ is not differentiable. Geometrically, the inequality in the definition of convexity means that if $P$, $Q$, and $R$ are any three points on the graph of $f$ with $Q$ between $P$, and $R$, then $Q$ is on or below the chord $PR$, or in terms of slopes \begin{equation} \text{slope }PQ\leq\text{slope }PR\leq\text{slope }QR, \label{1cvx slope inequality}% \end{equation} Now consider four points $w<x<y<z$ in $\mathbb{R}$ with, $P$, $Q$, $R$, $S$ the corresponding points on the graph of $f$. By the previous inequality \begin{equation} \text{slope }PQ\leq\text{slope }PR\leq\text{slope }QR\leq\text{slope }% QS\leq\text{slope }RS, \label{1cvx slope inequality two}% \end{equation} with strict inequalities if $f$ is strictly convex. Since slope $PR\leq$slope $QR$, we have that slope $QR$ increases as $x\nearrow y$, while slope $RS$ decreases as $z\searrow y$. Thus the left-hand side of the inequality $$ \frac{f\left( x\right) -f\left( y\right) }{x-y}\leq\frac{f\left( z\right) -f\left( y\right) }{z-y}% $$ increases as $x\nearrow y$ and the right-hand side decreases as $z\searrow y$. So there exist the left and right derivatives $f_{-}^{\prime}\left( y\right) \le f_{+}^{\prime}\left( y\right) $. Now take $m\in\left[ f_{-}^{\prime}\left( y\right) ,f_{+}^{\prime}\left( y\right) \right] $. Then $$m\leq f_{+}^{\prime}\left( y\right) \leq\frac{f\left( x\right) -f\left( y\right) }{x-y}\quad\text{if }x>y, $$ while $$\frac{f\left( x\right) -f\left( y\right) }{x-y}\leq f_{-}^{\prime }\left( x_{0}\right) \leq m\quad\text{if }x<y. $$ Hence, $f\left( x\right) -f\left( y\right) \geq m\left( x-y\right) $ for all $x$.