An integral inequality for variance Suppose the real valued function $f$ is differentiable on $[0,\,1]$ with bounded derivative. How do we prove the following? 
$$\int_0^1 f^2-\bigg(\int_0^1 f\bigg)^2\le \frac1{12}\sup_{x\in[0,1]}|f'(x)|^2.$$
I tried substituting $f(x)^2=f(0)^2+2\int_0^x f(t)f'(t)\,dt$, but it did not seem to work.
 A: We have
\begin{align*}
\left(f(x)-\int_{0}^{1}f(t)dt\right)^{2}&=\left(\int_{0}^{1}(f(x)-f(t))dt\right)^{2}\\
&=\left(\int_{0}^{1}(x-t)f'(\xi_{x,t})dt\right)^{2}\\
&\leq\|(f')^{2}\|_{\infty}\left(\int_{0}^{1}(x-t)dt\right)^{2},
\end{align*}
now 
\begin{align*}
\int_{0}^{1}\left(\int_{0}^{1}(x-t)dt\right)^{2}dx=\dfrac{1}{12},
\end{align*}
and note that $\text{Var}(X)=E(X-E(X))^{2}$.
Here I have used a version of Mean Value Theorem:
\begin{align*}
\int_{a}^{b}u(x)v(x)dx=c\int_{a}^{b}u(x)dx,
\end{align*}
where $m:=\inf v\leq c\leq M:=\sup v$, so in this context, 
\begin{align*}
\int_{0}^{1}(x-t)f'(\xi_{x,t})dt=c\int_{0}^{1}(x-t)dt,
\end{align*}
where $-\sup|f'|=\inf(-|f'|)\leq\inf f'\leq c\leq\sup f'\leq\sup|f'|$, so $c^{2}\leq(\sup|f'|)^{2}=\sup|f'|^{2}:=\|(f')^{2}\|_{\infty}$.
A: Let $X \sim U(0,1)$, then we may write the left-hand side as $\mathbb{V}(f(X))$. Now the variance is invariant under translations, so $\mathbb{V}(f(X)) = \mathbb{V}(f(X)-f(1/2))$. This yields
\begin{align}\mathbb{V}(f(X)-f(1/2)) &= \int_{0}^{1}(f(x)-f(1/2))^2 dx - \left(\int_{0}^{1} f(x) - f(1/2)dx\right)^2 \\
&\leq \int_{0}^{1} (f'(\xi_x)(x-1/2))^2 dx \leq \sup_{x\in [0,1]}|f'(x)|^2\int_{0}^{1}(x-1/2)^2 dx \\
&= \frac{1}{12}\sup_{x\in [0,1]}|f'(x)|^2 \\
&   \end{align} 
