If $f$ is a real continuously differentiable function defined on $[0,b]$ then $\int f^2\leq\int f'^2$ 
Suppose $f$ is a continuously differentiable real valued function defined on an interval $[0,b]$ with $f(0)=f(b)$. Prove that$$\int_0^bf'(x)^2dx\geq\int_0^bf(x)^2dx$$

I do have some doubts about this because if I consider $f\equiv C$ where $C$ is a constant then the inequality is false. So maybe it holds for non-constant functions only. I do not doubt the authenticity of this problem.
I am a bit stumped on how I should proceed. Please provide just a hint only to help me start. Thanks.

Suppose that $f(a)=f(b)=0$. 
We have $$f(x) = \int_a^x f'(s)ds = - \int_x^bf'(s)ds,$$hence
$$|f(x)|\le \min\left(\int_a^x |f'(s)|ds,\int_x^b|f'(s)|ds\right)\le \frac 12 \int_a^b|f'(s)|ds.$$
We integrate with respect to $x$ to get
$$\int_a^b f^2(x)dx\le \frac 14 (b-a)\left(\int_a^b|f'(s)|ds\right)^2\le \frac{(b-a)^2}{4}\int_a^b|f'(s)|^2ds$$
By Cauchy-Bunyakowski-Shwartz inequality.
Why the multiplicative constant $\frac{(b-a)^2}{4}$? To think of it, we can not avoid having some information on the domain of integration: its measure has to come into play.
Note that by similar technique we can get the result with the only hypothesis $f(a)=0$. We will lose the factor $\frac 14$, however. 
