Upper bound on $|\langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle|$ Let $H$ be a Hilbert space and $a,b,x$ be vectors in $H$.
I'm looking for a sharp upper bound on $|\langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle|$.
By the triangle inequality and Cauchy Schwarz this is clearly bounded by $2\|a\|\|b\|\|x\|^2$.
I think this is quite brutal, is there a bound with a constant less than $2$ ?
I tried rewriting the quantity as $\langle \langle a,x \rangle b - \langle a,b \rangle x, x \rangle$ or $\langle \langle a,x \rangle x - \langle x,x \rangle a, b \rangle$ but I can't get an improved upper bound this way.
 A: 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 $a$.
A: The span of the three vectors $\,a,b,x\,$ generate a subspace of at most
three dimensions. Thus, without loss of generality we can assume that they
are in $\,\mathbb R^3.\,$
Use the Binet-Cauchy identity
$$(a \times b)\cdot(c \times d) = (a \cdot c)(b \cdot d)-(a \cdot d)(b \cdot c) \tag1 $$
specialized to
$$(a \times x)\cdot(b \times x)=(a \cdot b)(x \cdot x)-(a \cdot x)(x \cdot b) \tag2 $$
and using $\,(a \cdot x) = (x \cdot a)\,$ to get
$$ \lVert(a \times x)\rVert^2 = (a \times x)\cdot(a \times x)=
(a \cdot a)(x \cdot x)-(a \cdot x)(x \cdot a) \le 
\lVert a\rVert^2 \lVert x\rVert^2 \tag3 $$
which implies
$$ \lVert(a \times x)\rVert \le \lVert a\rVert \lVert x\rVert \qquad
\text{ and } \qquad
\lVert(b \times x)\rVert \le \lVert b\rVert \lVert x\rVert. \tag4$$
Rearrange the terms of equation $(2)$ to get
$$ (a \cdot x)(x \cdot b)-(a \cdot b)(x \cdot x)=-(a \times x)\cdot(b \times x) \tag5. $$
Change notation to
$$ \langle a,x\rangle \langle x,b\rangle  - \langle a,b\rangle \langle x,x\rangle 
= -(a \times x)\cdot(b \times x). \tag6 $$
Use the results of equation $(4)$ on the right side and
$\, \langle x,b\rangle = \langle b,x\rangle \,$ to get
$$ |\langle a,x\rangle \langle b,x\rangle -\langle a,b\rangle \langle x,x\rangle| 
 \le \lVert a \rVert \lVert b \rVert \lVert x \rVert^2 \tag7. $$
A: We may assume without loss of generality that $a,b,x\in\mathbb{R}^3$ have unit modulus and
$$ x=(1,0,0)\qquad a=(\cos\theta,\sin\theta,0),\qquad b=(\cos\lambda\cos\phi,\sin\lambda\cos\phi,\sin\phi).  $$
Here we have $\langle a,x\rangle=\cos\theta$, $\langle b,x\rangle=\cos\lambda\cos\phi$ and $\langle a,b\rangle = \cos\theta\cos\lambda\cos\phi+\sin\theta\sin\lambda\cos\phi$, so our purpose is to find
$$ \max_{\theta,\lambda,\phi}\left|\cos\phi\right|\cdot\left|\cos\theta\cos\lambda-\cos(\theta-\lambda)\right|=\max_{\theta,\lambda}\left|\cos\theta\cos\lambda-\cos(\theta-\lambda)\right|. $$
The stationary points of the function $f(\theta,\lambda)=\cos\theta\cos\lambda-\cos(\theta-\lambda)$ occur at the points such that
$$ -\sin\theta\cos\lambda+\sin(\theta-\lambda) = 0 = -\cos\theta\sin\lambda-\sin(\theta-\lambda)$$
in particular along the curves described by $\sin(\theta+\lambda)=0$. Since
$$f(\theta,\pi n-\theta)=(-1)^n\left(\cos^2\theta-\cos(2\theta)\right)=(-1)^n\sin^2\theta$$
we have that
$$ \left|\langle a,x\rangle \langle b,x\rangle  - \langle a,b\rangle \langle x,x\rangle \right| \leq |a||b||x|^2.$$
The condition $|\cos\phi|=1$ is equivalent to the fact that $a,b,x$ lie on the same plane through the origin. If that is not the case the RHS can be further improved.

Alternative approach. This proves just a weaker inequality, but I want to outline it nevertheless since I promised in the comments. Assuming $a,b,x\in\mathbb{R}^n$ (even though it is sufficient to tackle the case $n=3$) we want to bound the absolute value of
$$ \sum a_j x_j \sum b_k x_k - \sum x_j^2 \sum a_k b_k = \sum_{j\neq k} x_j b_k(a_j x_k-x_j a_k) $$
where by Lagrange's identity
$$ \sum_{1\leq j < k\leq n}(a_j x_k-x_j a_k)^2 = \sum a_j^2 \sum x_k^2-\left(\sum a_j x_j\right)^2$$
such that
$$\left|\sum_{j\neq k}x_j b_k(a_j x_k-x_j a_k)\right|\leq \sqrt{\sum_{j\neq k}x_j^2 b_k^2\sum_{j\neq k}(a_j x_k-x_j a_k)^2}$$
where the RHS equals
$$ \sqrt{\left(\sum x_j^2\sum b_k^2-\sum x_j^2 b_j^2\right)\cdot 2\left(\sum x_j^2\sum a_k^2-\left(\sum a_j x_j\right)^2\right)} $$
such that $$ \left|\langle a,x\rangle \langle b,x\rangle  - \langle a,b\rangle \langle x,x\rangle \right| \leq \color{blue}{\sqrt{2}} |a||b||x|^2.$$
A: $
\newcommand\form[1]{\langle#1\rangle}
\renewcommand\Re{\operatorname{Re}}
\renewcommand\Im{\operatorname{Im}}
\newcommand\conj\bar
\newcommand\Ext{{\bigwedge}}
$
The inner product $\form{\cdot,\cdot}$ extends naturally to an inner product on the exterior algebra $\Ext H$ via
$$
  \form{v_1\wedge\cdots\wedge v_k,\:
        w_1\wedge\cdots\wedge w_l}
    = \delta_{kl}\det\Bigl(\form{v_i,v_j}\Bigr)_{i,j=1}^k,
\tag{$*$}
$$
i.e. take the determinant of the matrix with those entries. To see this, note that $\form{v,w}' := \form{v,\conj w}$ is a bilinear form; then is is well known that this extends naturally to a bilinear form $\form{\cdot,\cdot}'$ on $\Ext H$ via ($*$) but using $\form{\cdot,\cdot}'$, and we define $\form{X, Y} := \form{X, \conj Y}'$ for $X, Y \in \Ext H$. It is the easy to determine that ($*$) is satisfied.
Since $\form{\cdot,\cdot}$ is an inner product, it satisfies Cauchy-Schwarz. We see
$$
  |\form{x,x}\form{a,b} - \form{a,x}\form{b,x}|
    = |\form{x\wedge a, x\wedge b}|
    \leq \sqrt{\form{x\wedge a, x\wedge a}\form{x\wedge b, x\wedge b}}.
$$
But
$$
  \form{x\wedge a, x\wedge a}
    = |x|^2|a|^2 - \form{x, a}\form{a, x}
    = |x|^2|a|^2 - |\form{x, a}|^2
    \leq |x|^2|a|^2,
$$
and the same for $\form{x\wedge b, x\wedge b}$. Hence
$$
  |\form{x,x}\form{a,b} - \form{a,x}\form{b,x}| \leq |x|^2|a||b|.
$$
A: Assume WLOG, $\left\langle a, a\right\rangle = 1$. Let $A = \frac12 \left(ab^T + ba^T\right)$, $t = \left\langle a, b\right\rangle$ and $c = b - ta$. Then $\left\langle c, a\right\rangle = 0$, $\left\langle b, b\right\rangle = t^2 + \left\langle c, c \right\rangle$ and $$A = taa^T + \frac{1}{2}\left(ac^T + ca^T\right).$$
Let's find the set of eigenvalues of $A$. Since $\text{rank}(A) \le 2$ ($\text{Im}(A) \subset \text{Vect}(a, c)$), it has at most two non-zero eigenvalues. Since $$Aa = ta + \frac{1}{2} c$$ and $$Ac = \frac{1}{2}\left\langle c, c \right\rangle a,$$ the non-zero eignevalues (if they exist) are solutions of the equation:
$$\mu^2 - t\mu - \frac{1}{4}\left\langle c, c\right\rangle$$
This proves that:
$$\lambda \left(A\right) = \left\{0, \frac{t \pm \sqrt{t^2 + \left\langle c, c\right\rangle}}{2}\right\} = \left\{0, \frac{\left\langle a, b\right\rangle \pm \left\|b\right\|}{2}\right\}.$$
Now back to the question:
$$\left|\left\langle a, x\right\rangle\left\langle b, x\right\rangle - \left\langle a, b\right\rangle\left\langle x, x\right\rangle \right| = \left|\left\langle x, \left(A - \left\langle a, b\right\rangle I\right)x\right\rangle\right| \le \left(\max\limits_{\mu \in \lambda\left(A - \left\langle a, b\right\rangle I\right)} |\mu|\right)\left\|x\right\|^2 $$
However, \begin{align}
\max\limits_{\mu \in \lambda\left(A - \left\langle a, b\right\rangle I\right)} |\mu| &= \max\limits_{\mu \in \lambda\left(A\right)} |\mu - \left\langle a, b\right\rangle|\\
&= \max \left\{\left|\left\langle a, b\right\rangle\right|, \left| \frac{\left\langle a, b\right\rangle \pm \left\|b\right\|}{2}-\left\langle a, b\right\rangle \right|\right\}\\
&= \max \left\{\left|\left\langle a, b\right\rangle\right|, \left|\frac{\left\|b\right\| + \left\langle a, b\right\rangle}{2}\right|, \left|\frac{\left\|b\right\| - \left\langle a, b\right\rangle}{2}\right|\right\} \le \frac{1}{2}\left(\left\|b\right\| + \left|\left\langle a, b\right\rangle \right|\right).
\end{align}
And you will find the better inequality:
$$\left|\left\langle a, x\right\rangle\left\langle b, x\right\rangle - \left\langle a, b\right\rangle\left\langle x, x\right\rangle \right| \le \frac{1}{2}\left(\left\|a\right\|\left\|b\right\| + \left|\left\langle a, b\right\rangle\right|\right)\left\|x\right\|^2$$
