Show that $\max_{x\in[a,b]}|f'(x)|\geq\frac{4}{(b-a)^2}\int_a^b|f(x)|dx$ Let $f:[a,b]\to\mathbb{R}$ be differentiable and $f'$ is continuous. Suppose $f(a)=f(b)=0$, show that
$$\max_{x\in[a,b]}|f'(x)|\geq\frac{4}{(b-a)^2}\int_a^b|f(x)|dx$$
My approach. For any $x$, by Taylor's theorem there is $\xi\in(a,x)$ s.t.
$$|f(x)|=|f(a)+f'(\xi)(x-a)|\leq\max_{x\in[a,b]}|f'(x)|(x-a)$$
Hence 
$$\int_a^b |f(x)|dx\leq\max_{x\in[a,b]}|f'(x)|\int_a^bx-adx=\frac{(b-a)^2}{2}\max_{x\in[a,b]}|f'(x)|$$
I can only get this, and I don't know how to obtain $\frac{(b-a)^2}{4}$.

Where do you use that $f(b)=0?$ why not try running the argument from both ends?! 
EDIT
OK, note that $$|f(x)| \leq \max |f^\prime(x)| (\min(x-a, b-x)).$$
Now integrate both sides.
