$f$ is convex in $\mathbb{R}$ then $f'$ is strictly increasing in $\mathbb{R}$
If $x<x_0$
$f$ continuous in $[x,x_0]$ and differentiable in $(x,x_0)$, from the Mean Value Theorem
\begin{align*} \exists &\xi \in (x,x_0):f'(\xi)=\frac{f(x_0)-f(x)}{x_0-x}=\frac{f(x)-f(x_0)}{x-x_0} \\ x_0>\xi \Rightarrow f'(x_0)&>f'(\xi) \\ \Rightarrow f'(x_0)&>\frac{f(x)-f(x_0)}{x-x_0} \\ \Rightarrow f'(x_0)(x-x_0)&<f(x)-f(x_0) \\ \Rightarrow f(x)&>f'(x_0)(x-x_0)+f(x_0) \end{align*} If $x>x_0$
$f$ continuous in $[x_0,x]$ and differentiable in $(x_0,x)$, from the Mean Value Theorem
\begin{align*} \exists & \xi \in (x_0,x):f'(\xi)=\frac{f(x)-f(x_0)}{x-x_0} \\ x_0<\xi \Rightarrow f'(x_0)&<f'(\xi) \\ \Rightarrow f'(x_0)&<\frac{f(x)-f(x_0)}{x-x_0} \\ \Rightarrow f'(x_0)(x-x_0)&<f(x)-f(x_0) \\ \Rightarrow f(x)&>f'(x_0)(x-x_0)+f(x_0) \end{align*}
Edit:
If $x=x_0$ then $f(x_0)\geq f(x_0) $ which holds as an equality.
In the end $f(x)\geq f'(x_0)(x-x_0)+f(x_0) \, \forall x \in \mathbb{R}$