Graph of convex function $f$ lies above line that passes through point $(x, f(x))$? 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 $y \in \mathbb{R}$?
 A: 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 $x$.  
