Here is a upper bound:
Assume that $n \ge 3$.
Denote
$S = \{(v_1, v_2, \cdots, v_n) : ~ v_i \ge 0, \forall i; ~ v_1 + v_2 + \cdots + v_n = 1\}.$
We have
$$\max_{(v_1, v_2, \cdots, v_n) \in S} \frac{\sum_i v_i^3}{(\sum_i v_i^2)^2} \le \frac{3\sqrt 3}{16}\sqrt n + \frac58 + \frac{5\sqrt 3}{96\sqrt n}. \tag{1}$$
The bound (1) satisfies that
$$\lim_{n\to \infty} \left(\frac{3\sqrt 3}{16}\sqrt n + \frac58 + \frac{5\sqrt 3}{96\sqrt n} - \max_{(v_1, v_2, \cdots, v_n) \in S} \frac{\sum_i v_i^3}{(\sum_i v_i^2)^2}\right) = 0. \tag{2}$$
Proof of (1) and (2):
Consider the maximum of
$$f(v_1, v_2, \cdots, v_n) = \frac{\sum_i v_i^3}{(\sum_i v_i^2)^2}$$
subject to $v_i \ge 0, \forall i; ~ \sum_{i=1}^n v_i = 1$.
Using Vasc's Equal Variable Theorem (Corollary 1.9, [1]), $f$
is maximal when $0 \le v_1 = v_2 = \cdots = v_{n - 1} \le v_n$.
Letting $v_n = x$ and $v_1 = v_2 = \cdots = v_{n-1} = \frac{1 - x}{n-1}$, we have
$$\max_{(v_1, v_2, \cdots, v_n) \in S} \frac{\sum_i v_i^3}{(\sum_i v_i^2)^2} = \max_{x\in [0, 1]} \frac{(n^2 - 2n)x^3 + 3x^2 - 3x + 1}{(nx^2 - 2x + 1)^2}.$$
It suffices to prove that, for all $x\in [0, 1]$ and $n \ge 3$,
$$\frac{(n^2 - 2n)x^3 + 3x^2 - 3x + 1}{(nx^2 - 2x + 1)^2} \le \frac{3\sqrt 3}{16}\sqrt n + \frac58 + \frac{5\sqrt 3}{96\sqrt n}.$$
Letting $m = \sqrt{3n} - 3\ge 0$. After clearing the denominators, it suffices to prove that, for all $x \in \mathbb{R}$,
$$a_1 x^4 + b_1x^3 + c_1x^2 + d_1 + e_1 \ge 0 \tag{3}$$
where
\begin{align*}
a_1 &= 6\,{m}^{6}+128\,{m}^{5}+1115\,{m}^{4}+5100\,{m}^{3}+12960\,{m}^{2}+
17388\,m+9639, \\
b_1 &= -32\,{m}^{5}-552\,{m}^{4}-3792\,{m}^{3}-13020\,{m}^{2}-22392\,m-15444, \\
c_1 &= 36\,{m}^{4}+552\,{m}^{3}+3270\,{m}^{2}+8460\,m+8118,\\
d_1 &= -216\,{m}^{2}-1152\,m-1692, \\
e_1 &= 54\,{m}^{2}+216\,m+207.
\end{align*}
We need the following auxiliary result.
Fact 1: Let $a_1 > 0, b_1, c_1, d_1, e_1$ be real numbers.
Let $f(x) = a_1x^4 + b_1x^3 + c_1x^2 + d_1x + e_1$.
Let $\Delta_f$ denote its discriminant.
Let
$D_f = 64a_1^3e_1-16a_1^2c_1^2+16a_1b_1^2c_1
-16a_1^2b_1d_1-3b_1^4$.
If $\Delta_f \ge 0$ and $D_f > 0$, then $f(x)\ge 0$ for any real number $x$.
(See https://en.wikipedia.org/wiki/Quartic_function and the reference [13] therein,
i.e., Rees, E. L. (1922). "Graphical Discussion of the Roots of a Quartic Equation". The American Mathematical Monthly. 29 (2): 51–55.)
We have
\begin{align*}
\Delta_f &= 63700992\, \left( 6\,{m}^{2}+24\,m+23 \right)\\
&\qquad \times \left( 512\,{m}^{5}+
5886\,{m}^{4}+26604\,{m}^{3}+57887\,{m}^{2}+58242\,m+19326 \right) \\
&\qquad \times
\left( {m}^{2}+6\,m+6 \right) ^{4} \left( m+3 \right) ^{6}\\
&> 0
\end{align*}
and
\begin{align*}
D_f &= 12288\, \left( 4\,{m}^{6}+96\,{m}^{5}+916\,{m}^{4}+4460\,{m}^{3}+11389
\,{m}^{2}+13734\,m+5262 \right) \\
&\qquad \times \left( 8\,{m}^{5}+138\,{m}^{4}+900\,{
m}^{3}+2769\,{m}^{2}+4086\,m+2358 \right) \left( m+3 \right) ^{9}\\
&> 0.
\end{align*}
By Fact 1, (3) is true.
(2) is true since
$$\lim_{n\to \infty} \left(\frac{3\sqrt 3}{16}\sqrt n + \frac58 + \frac{5\sqrt 3}{96\sqrt n} - f\Big(\sqrt{3/n}\Big)\right) = 0$$
where
$$f(x) := \frac{(n^2 - 2n)x^3 + 3x^2 - 3x + 1}{(nx^2 - 2x + 1)^2}.$$
We are done.
Reference:
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007.
Online: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
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 $\sup_{\substack{\|x\|=\|y\|=1}}\langle y, Tx\rangle \leq \sup_{\|x\|=1}\|Tx\| = \|T\|$.
Best Answer
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 $a$.