Let's prove that $\lvert \langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle \lvert \le \lVert a \rVert \lVert b \rVert \lVert x \rVert^2$.
Without loss of generality, we can suppose that $\lVert a \rVert = \lVert b \rVert = \lVert x \rVert=1$. Now take an orthonormal basis $\{e_1, e_2, \dots\}$ of $H$ such that $a= \cos \theta e_1 + \sin \theta e_2$ and $b= \cos \theta e_1 - \sin \theta e_2$, where $0 \le \theta \le \frac{\pi}{2}$. We get
$$\begin{aligned}
\langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle &= (\cos \theta \langle e_1,x \rangle + \sin \theta \langle e_2,x \rangle)(\cos \theta \langle e_1,x \rangle - \sin \theta \langle e_2,x \rangle) - \cos 2\theta\\
&=\cos^2 \theta\langle e_1,x \rangle^2 - \sin^2 \theta \langle e_2,x \rangle^2 - \cos 2 \theta\\
&=(\langle e_1,x \rangle^2 + \langle e_2,x \rangle^2) \cos^2 \theta - \langle e_2,x \rangle^2 - \cos 2 \theta\\
&=\frac{\langle e_1,x \rangle^2 - \langle e_2,x \rangle^2}{2}-\left(1 - \frac{\langle e_1,x \rangle^2 + \langle e_2,x \rangle^2}{2}\right) \cos 2 \theta
\end{aligned}$$
Even if it means swapping $a$ and $b$, we can suppose that $\lvert \langle e_1,x \rangle \rvert \ge \lvert \langle e_2,x \rangle \rvert$. The quantity above depends on $\theta$ and is positive and maximum for $\theta = \pi$, i.e. $b=-a$. In that case
$$\begin{aligned}
\lvert \langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle \lvert &\le 1 - \langle e_2,x \rangle^2 \le 1
\end{aligned}$$ and we get the inequality
$$\lvert \langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle \lvert \le \lVert a \rVert \lVert b \rVert \lVert x \rVert^2$$ which is an equality if and only if $b= \pm a$ and $x$ is orthogonal to $a$.
Best Answer
It is not necessarily true. Consider next example: let $H = \mathbb{R}^2$ with the dot product and $T(x_1, x_2) = (-x_2, x_1)$, then $\langle x, Tx\rangle = (x_1, x_2) \cdot (-x_2, x_1) = 0 \;\forall (x_1, x_2) \in \mathbb{R}^2$. Obviously $T$ is bounded operator with norm $\|T\| = \sup_{\|x\|=1} \|Tx\| = \sup_{\|x\|=1} \|x\|=1$ (since $\|Tx\|=\|x\|$).
However it is true that $\|T\| = \sup_{\substack{\|x\|=1, \|y\|=1}}\langle y, Tx\rangle$. This is obvious when $T \equiv 0$, overwise, we can find $x$, such that $\|x\|=1$ and $Tx \neq 0$. Take for them $y := \frac{Tx}{\|Tx\|}$, then $\|y\|=1$, therefore $$\sup_{\|y\|=\|x\|=1}\langle y, Tx\rangle = \sup_{\|y\|=\|x\|=1, Tx \neq 0}\langle y, Tx\rangle \geq \sup_{\|x\|=1, Tx \neq 0} \frac{\langle Tx, Tx\rangle}{\|Tx\|} = \sup_{\|x\|=1, Tx \neq 0} \|Tx\| = \|T\|$$
From the Cauchy inequality we know that $|\langle y, Tx\rangle| \leq \|y\| \cdot \|Tx\|$, therefore $\sup_{\substack{\|x\|=\|y\|=1}}\langle y, Tx\rangle \leq \sup_{\|x\|=1}\|Tx\| = \|T\|$.