Let $\mu$ be the standard Gaussian distribution on $\mathbb{R}$. Show that if $B$ is a Borel set (w.r.t the Euclidean metric) and $\mu(B)\geq 1/2$ then
$$\mu(B_{r})\geq 1-\frac{1}{2}e^{-\frac{t^2}{2}} $$
For all $r>0$, where $B_{r} := \{x \in \mathbb{R} : \inf_{y\in B} |x-y|<r \}$.
Attempt : Since $\mu(B)\geq 1/2$ (and since we are centred) it follows $\mu(B_{r})\geq \mu((-\infty,r])$
So $$\mu(B_{r})\geq 1- \int_{r}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-(t^2)/2}dt\geq 1-\frac{1}{\sqrt{2\pi}r}e^{-(r^2)/2}$$