TITLE: inequality for positive contraction operator
QUESTION [6 upvotes]: Let $H$ be a Hilbert space, let $A\in B(H)$ satify $\|A\|\le 1$. If $A$ is positive, i.e. $A$ is a self-adjoint operator and for all $x\in H$, $\langle A(x),x\rangle\ge 0$, proof that 
$${\|x-A(x)\|}^2\le {\|x\|}^2-{\|A(x)\|}^2, \forall x\in H.$$
This is a exercise. For $A$ is positive
\begin{align*}
{\|x-A(x)\|}^2
&={\|x\|}^2+{\|A(x)\|}^2-\langle A(x),x\rangle-\langle x,A(x)\rangle\\
&={\|x\|}^2-{\|A(x)\|}^2+2{\|A(x)\|}^2-2\langle x,A(x)\rangle
\end{align*}
so we should prove that 
$${\|A(x)\|}^2\le \langle x,A(x)\rangle$$
or 
$${\|A(x)\|}^2=\langle A(x),A(x)\rangle = \langle x,A(A(x))\rangle \le \langle x,A(x)\rangle$$
but i don't know how to use the condition "$\|A\|\le 1$" and "$\langle A(x),x\rangle\ge 0$".

REPLY [5 votes]: 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 $\langle Ax,x\rangle \ge \|Ax\|^2$.