# inequality for positive contraction operator

### Solution:

Since $$A$$ is a positive operator, you can take it square root, which is also a positive operator, and $$\|A(x)\|^2 = \langle A^{1/2}A^{1/2}x,A x\rangle = \langle A^{1/2}x,A (A^{1/2}x)\rangle$$ now by Cauchy-Schwarz and the fact that $$\|A\|\leq 1$$ $$\langle A^{1/2}x,A (A^{1/2}x)\rangle ≤ \|A^{1/2}(x)\|\|A(A^{1/2}(x))\| ≤ \|A^{1/2}(x)\|^2 = \langle x, A(x)\rangle$$ which is what you wanted to prove.

For an alternative solution which doesn't involve the square root, consider this. From $$\|A\| \le 1$$ it follows that for all $$x \in H$$ holds $$\langle Ax,x\rangle \le \|Ax\|\|x\| \le \|x\|^2.$$ Therefore \begin{align} 0 &\le \langle A(Ax-x),Ax-x\rangle \\ &= \langle A^2x - Ax, Ax - x\rangle\\ &= \underbrace{\langle A^2x,Ax\rangle}_{\le\|Ax\|^2} - \underbrace{\langle A^2x,x\rangle}_{=\|Ax\|^2} - \underbrace{\langle Ax,Ax\rangle}_{=\|Ax\|^2} + \langle Ax,x\rangle\\ &\le \langle Ax,x\rangle - \|Ax\|^2 \end{align}

so $$\langle Ax,x\rangle \ge \|Ax\|^2$$.