Distance to a closed subspace of a Hilbert space in terms of inner product with the unit normal

Question

Let $M$ be a closed subspace of a Hilbert space $H$, and suppose $x_0\in H$ Show that:

$$\min(\|m-x_0\|, m\in M)=\max(|\langle x_0,n\rangle|, n\in M^\perp ,\|n\|=1)$$

I know that

$|\langle x,y\rangle|^2 \leq \langle x,x\rangle\langle y,y\rangle \implies |\langle x,y\rangle|\leq \sqrt{\langle x,x\rangle}\sqrt{\langle y,y\rangle}\implies \|x\| \|y\|$

But I'm not sure where else to go from here...