Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.
For $A\in \mathcal{B}(F)$ we set $$n(T)=\sup\Big\{\big|\langle A y,y\rangle\big|\;:y \in F,\;\;\|y\|=1\Big \}.$$
I want to prove that $$n(T)=\inf\Big\{\lambda>0:\;\big|\langle A y,y\rangle\big|\leq\lambda\|y\|^2,\;\forall\,y\in F\Big\}.$$
Let $y \in F$ be such that $y\neq 0$. Then $$\big|\langle A y,y\rangle\big|=\left|\left\langle A \frac{y}{\|y\|},\frac{y}{\|y\|}\right\rangle^{\vphantom{a^a}}\right| \,\|y\|^2\leq n(T)\|y\|^2.$$ So, $$\big|\langle A y,y\rangle\big|\leq n(T)\|y\|^2.$$ If $y=0$, then the above inequality holds and so $$\big|\langle A y,y\rangle\big|\leq n(T)\|y\|^2,$$ for all $y\in F$. Hence $$\inf\Big\{\lambda>0:\;\big|\langle A y,y\rangle\big|\leq\lambda\|y\|^2,\;\forall\,y\in F\Big\}\leq n(T).$$ I'm facing difficulties to prove that $$\inf\Big\{\lambda>0:\;\big|\langle A y,y\rangle\big|\leq\lambda\|y\|^2,\;\forall\,y\in F\Big\}\geq n(T).$$
Let $\ \epsilon>0\ $, and $ y_\epsilon\in F\ $, such that $\ \|y_\epsilon\|=1\ $ and $\ \left|\langle Ay_\epsilon,y_\epsilon\rangle\right|\ \ge n(T)-\epsilon\ $. Since every $\ \lambda\ $ in the set $\ \left\{\lambda>0\,:\, \left|\langle Ay,y\rangle\right|\le \lambda\|y\|^2, \forall y\in F\right\}\ $ must satisfy the inequalities \begin{align} \lambda&=\lambda\|y_\epsilon\|^2\\ &\ge \left|\langle Ay_\epsilon,y_\epsilon\rangle\right|\\ &\ge n(T)-\epsilon\ , \end{align} then $\ \inf\left\{\lambda>0\,:\, \left|\langle Ay,y\rangle\right|\le \lambda\|y\|^2, \forall y\in F\right\}\ge n(T)-\epsilon\ $ as well. Since $\ \epsilon\ $ was arbitrary, it follows that $$ \inf\left\{\lambda>0\,:\, \left|\langle Ay,y\rangle\right|\le \lambda\|y\|^2, \forall y\in F\right\}\ge n(T)\ . $$