Show that $\int\limits_a^b |f(t)|dt \leq (b-a)\int\limits_a^b|f'(t)|dt$ 
Let $f:[a,b]\to\mathbb{R}$ be continuously differentiable. Suppose $f(a) = 0$. Show that
  $$
\int\limits_a^b|f(t)|dt \leq (b-a)\int\limits_a^b|f'(t)|dt
$$

By the mean value theorem, for every $t\in[a,b]$, there exists $c\in[a,t]$ such that
$$
\frac{f(t) - f(a)}{t-a} = f'(c) \Rightarrow f(t) = (t-a)f'(c) \leq (b-a)f'(c)
$$
So basically, I have that $f(t) \leq (b-a)f'(c)$. But $c$ depends on $t$, so I don't know how to proceed from here.

I think the mean value theorem is a no-go here. If you start with (just the fundamental theorem of calculus)
$$
f(t)=f(a)+\int_a^t f'(s)\,ds=\int_a^t f'(s)\,ds,
$$
you find with the triangle inequality that
$$
|f(t)|\leq \int_a^t |f'(s)|\,ds.
$$
Next, integrate this inequality from $a$ to $b$:
$$
\int_a^b |f(t)|\,dt\leq \int_a^b\int_a^t |f'(s)|\,ds\,dt.
$$
In the final integral, change order of integration and estimate $(b-s)\leq(b-a)$. I leave that part to you.
