# Proving an inequality using Cauchy-Schwarz

I am supposed to use the Cauchy Schwarz inequality to solve this problem but I am stuck. I can't see how the square root of $N$ comes out. Could anyone please help me?

## 2 Answers

If $x_1,\ldots,x_N$ are a basis of orthonormal vectors and $\psi$ has a unit norm, $\psi$ can be decomposed as $$\psi = \sum_{k=1}^{N}\alpha_k x_k,\qquad\sum_{k=1}^{n}\alpha_k^2=1$$ and $$\begin{eqnarray*} \sum_{k=1}^{N}\|\psi-x_k\|^2 &=& \sum_{k=1}^{N}\left[(\alpha_k-1)^2+\sum_{j\neq k}\alpha_j^2\right]\\&=&\sum_{k=1}^{N}\left[2-2\alpha_k \right]=2N-2\sum_{k=1}^{N}\alpha_k\end{eqnarray*}$$ where by the Cauchy-Schwarz inequality $$\left|\sum_{k=1}^{N}\alpha_k\right|^2\leq \sum_{k=1}^{N}1\sum_{k=1}^{N}\alpha_k^2 = N,$$ hence: $$\sum_{k=1}^{N}\|\psi-x_k\|^2 \geq 2N-2\sqrt{N}$$ as wanted.

Observe that

$$\|\psi-x\|^{2}=\left<\psi-x,\psi-x\right>=\|\psi\|^{2}+\|x\|^2-\left<\psi,x\right>-\left<x,\psi\right>=2-2{\rm Re}\left<\psi,x\right>$$

and thus

$$\sum_{x}\|\psi-x\|^{2}=2N-2{\rm Re}\left<\psi,\sum_{x}x\right>$$

But

$${\rm Re}\left<\psi,\sum_{x}x\right>\leq\left|\left<\psi,\sum_{x}x\right>\right|\leq\|\psi\|\left\|\sum_{x}x\right\|=1\cdot\sqrt{N}$$

where the last equality follows from the fact that the $x$'s are orthogonal. This finishes the proof yielding

$$\sum_{x}\|\psi-x\|^{2}\geq 2N-2\sqrt{N}$$