In our introductory functional analysis course, we defined the norm of a bounded operator $T \in B(H)$ as $$ \|T\| := \sup\_{\| x \| \leq 1} \|T(x)\|. $$ 

However, there is now a lemma saying:

**Lemma:** Let $T \in B(H)$ be self-adjoint. Then $$ \| T\| = \sup \left\{ | \langle Tx, x \rangle | \ | x \in H, \| x\| \leq 1 \right\}. $$ 

For the proof, one part of the inequality I understand. We let $$ M = \sup \left\{ | \langle Tx, x \rangle | \ | x \in H, \| x\| \leq 1 \right\}. $$ Then I have that by Cauchy-Schwarz $$ | \langle Tx, x \rangle | \leq \| T(x)\| \ \| x \| \leq \| T \| \ \| x \|^2. $$ Taking this for all $x \in H$ with $\| x \| \leq 1$, we see that $M \leq \| T\|. $

However, in my course notes, in order to prove that $M \geq \|T\|$, the claim is being made that **the following holds**: $$ \pm \langle Tx, x \rangle \leq M \|x\|^2.$$ (Since $T$ is self-adjoint, $\langle Tx, x \rangle \in \mathbb{R}$ and this makes sense). 

One then uses this in the inequalities $$\langle T(x+y), x+y \rangle \leq M \|x+y\|^2 \\ - \langle T(x-y), x-y\rangle \leq M \|x-y\|^2$$ adds these to take the real part, and then one arrives at the desired conclusion. 

**My question** is, how to prove that $$ \pm \langle Tx, x \rangle \leq M \|x\|^2.$$