Let $d\mu$ be a Gaussian measure on $\mathbb{R}$ with the center $a \in \mathbb{R}$ and variance $1$.

Let $B(a,r) \subset \mathbb{R}$ be the interval $[a-r,a+r]$.

Then, for any smooth mapping $f : \mathbb{R} \to \mathbb{R}$, is it true that \begin{equation} \frac{1}{\mu(B(a,r))}\int_{B(a,r)} f^2(x) d\mu(x) - \Bigl( \frac{1}{\mu(B(a,r))}\int_{B(a,r)} f(x) d\mu(x) \Bigr)^2 \leq \ \int_{B(a,r)} [f'(x)] ^2 d\mu(x)? \end{equation} for any $r>0$?

Due to smoothness of $f$, we observe that both sides of the above inequality tends to zero as $r \to 0^+$.

I guess that due to concentration of the Gaussian measure around the center, this inequality must hold for any $r>0$, but I cannot really find a relelvant reference to help justify my guess.

Could anyone please help me?