Prove $g(x)=\frac{f(x)-f(a)}{x-a}$ is increasing 
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a differentiable function such that $f'$ is increasing. Setting $a\in\mathbb{R}$, prove: the function $g(x)=\frac{f(x)-f(a)}{x-a}$ is also an increasing function on $(a,\infty)$ and $(-\infty,a)$

My thoughts:
By MVT: $\exists c\in(a,x_0) \ s.t. f'(c)=\frac{f(x_0)-f(a)}{x-a}, \ \exists d\in(x_0,x_1) \ s.t. \ f'(d)=\frac{f(x_1)-f(x_0)}{x_1-x_0}$
$d>c \Rightarrow f'(d)\geq f'(c)$
Now I got stuck since I couldn't show $\frac{f(x_1)-f(a)}{x_1-a} \geq \frac{f(x_1)-f(x_0)}{x_1-x_0}$.
Any help appreciated.

Note that $g$ is differentiable when $x\ne a$. Thus, we only need to show that
$$g'(x)=\frac{f'(x)(x-a)-(f(x)-f(a))\cdot1}{(x-a)^2}\ge 0,$$
or
$$f'(x)(x-a)\ge f(x)-f(a).$$
If $x>a$, we only need to show that $$f'(x)\ge\frac{f(x)-f(a)}{x-a}.$$
By MVT, the RHS is $f'(c)$ for some $c\in[a,x]$. And clearly
$$f'(x)\ge f'(c).$$
