$a \le \|a\|$ in a von Neumann algebra

Question

Let $M$ be a von Neumann Algebra acting on a Hilbert space $H$. Let $a$ be a positive element of $M$. Then I want to show that $a \le \|a\|1$.
My attempt: Since $a\ge 0$, there exists an element $x \in M$ such that $a=x^*x$. I have to show that $\langle a \xi,\xi\rangle \le \langle \|a\|\xi,\xi\rangle$ for all $\xi \in H$. Now notice that, $$\langle a \xi,\xi\rangle=\langle x^*x \xi,\xi\rangle=\|x\xi\|^2.$$ Now I am unable to proceed from here. Please help me to solve this. Thank you for you time and contribution.