# Cauchy-Schwarz inequality for symmetric operators

Suppose that A is a symmetric operator such that $(Au,u)\geq 0$ where $(\cdot,\cdot)$ denotes the inner product.

How do I show that $|(Au,v)|\le (Au,u)^{1/2} (Av,v)^{1/2}$? I can't figure out how the Cauchy-Schwarz inequality is used in this case...

The easy way to solve the problem is to notice that $(A u,v) = (A^{1/2}u,A^{1/2}v)$. Another way is to mimic the proof of Cauchy-Schwarz, that is, start with the inequality $(Au + tAv,u+tv) \ge 0$, and then find the minimal value of the quantity on the left hand side (e.g. by differentiating and setting the derivative equal to zero).
Let $\epsilon > 0$. Then $(x,y)_{\epsilon}=((A+\epsilon I)x,y)$ is an inner-product because $(x,x)_{\epsilon}\ge 0$ with equality iff $x=0$. So, $$|(x,y)_{\epsilon}|^{2} \le (x,x)_{\epsilon}(y,y)_{\epsilon}.$$ In other words, $$|((A+\epsilon I)x,y)|^{2}\le ((A+\epsilon I)x,x)((A+\epsilon I)y,y),\;\;\; x,y \in X.$$ All of the expressions are continuous in $\epsilon$ for $\epsilon \ge 0$. Letting $\epsilon \rightarrow 0$ gives $$|(Ax,y)|^{2} \le (Ax,x)(Ay,y).$$