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.

For $a\le x\le b$, define
$$F(x)=(x-a)\int_a^x|f'(t)|dt-\int_a^x|f(t)|dt$$
Note that $F(a)=0$ and
$$F'(x)=\int_a^x|f'(t)|dt+(x-a)|f'(x)|-|f(x)|\ge\int_a^x|f'(t)|dt-|f(x)|$$
But, since $f(a)=0$, we have $f(x)=f(x)-f(a)=\int_1^xf'(t)dt$, hence
$$|f(x)|=\left|\int_a^xf'(t)dt\right|\le\int_a^x|f'(t)|dt$$
which implies $F'(x)\ge0$ for $a\le x\le b$, which, in turn, implies $F(b)\ge F(a)=0$, and this gives the desired result.
