Integral Inequality (possibly related to Green functions) Let $f\in C^1([0,1]) $ such that $f(0)=f(1)=0 $
Show that $$\int_0^1|f(x)|^2dx\leq \frac{1}{8}\int_0^1|f'(x)|^2dx $$
I'm not sure if this inequality has a special name, but I haven't been able to find anything on it. I tried decomposing $f$ on the left side into an arbitrary constant plus the integral of its derivative, but it just seemed to complicate things further. I also considered taking the square root of both sides but I didn't get very far with that either. I'm thinking maybe there's a more theoretical aspect to it (Rolle's/MVT)
This inequality reminds me a lot of the following relating the infinity norm of the Green function:
Given $u(x)=\int_0^1G(x,y)f(y)dy$
$$||u(x)||_{\infty}\leq\frac{1}{8}||f(x)||_{\infty} $$
Perhaps their proofs are similar?
 A: Since $f(x) = f(x) - f(0) = \int\limits_{0}^{x}{f'(t)\text{ d}t}$, we can apply Cauchy Schwarz to yield
$$|f(x)|^2 = \left(\int\limits_{0}^{x}{f'(t)\cdot 1\text{ d}t}\right)^2 \le\int\limits_{0}^{x}{|f'(t)|^2\text{ d}t}\int\limits_{0}^{x}{1\text{ d}t} = x\int\limits_{0}^{x}{|f'(t)|^2\text{ d}t}.$$
Similarly, since $f(x) = f(x) - f(1) = -\int\limits_{x}^{1}{f'(t)\text{ d}t}$, a similar calculation yields
$$|f(x)|^2 = \left(\int\limits_{x}^{1}{f'(t)\cdot 1\text{ d}t}\right)^2 \le\int\limits_{x}^{1}{|f'(t)|^2\text{ d}t}\int\limits_{x}^{1}{1\text{ d}t} = (1-x)\int\limits_{x}^{1}{|f'(t)|^2\text{ d}t}.$$
Hence, we have
\begin{align}
\int\limits_{0}^{1}{|f(x)|^2\text{ d}t} &= \int\limits_{0}^{1/2}{|f(x)|^2\text{ d}x} + \int\limits_{1/2}^{1}{|f(x)|^2\text{ d}x} \\
&\le\int\limits_{0}^{1/2}{x\int\limits_{0}^{x}{|f'(t)|^2\text{ d}t}\text{ d}x} + \int\limits_{1/2}^{1}{(1-x)\int\limits_{x}^{1}{|f'(t)|^2\text{ d}t}\text{ d}x} \\
&\le\int\limits_{0}^{1/2}{x\int\limits_{0}^{1/2}{|f'(t)|^2\text{ d}t}\text{ d}x} + \int\limits_{1/2}^{1}{(1-x)\int\limits_{1/2}^{1}{|f'(t)|^2\text{ d}t}\text{ d}x} \\
&=\left(\int\limits_{0}^{1/2}{x\text{ d}x}\right)\left(\int\limits_{0}^{1/2}{|f'(t)|^2\text{ d}t}\right) + \left(\int\limits_{1/2}^{1}{(1-x)\text{ d}x}\right)\left(\int\limits_{1/2}^{1}{|f'(t)|^2\text{ d}t}\right) \\
&=\frac{1}{8}\left(\int\limits_{0}^{1/2}{|f'(t)|^2\text{ d}t}+\int\limits_{1/2}^{1}{|f'(t)|^2\text{ d}t}\right) \\
&=\frac{1}{8}\int\limits_{0}^{1}{|f'(t)|^2\text{ d}t}
\end{align}
as desired.
