In [this great question](https://mathoverflow.net/q/301844/102946) by Nathaniel Johnston, and in its answers, we can learn the following remarkable inequality: For all $v,w \in \mathbb{R}^n$ we have
\begin{align\*}
 \|v^2\| \, \|w^2\| - \langle v^2, w^2 \rangle \le \|v\|^2 \|w\|^2 - \langle v,w \rangle^2; \quad (\*)
\end{align\*}
here, $\langle\cdot,\cdot\rangle$ denotes the standard inner product on $\mathbb{R}^n$, $\|\cdot\|$ denotes the Euclidean norm and $v^2,w^2 \in \mathbb{R}^n$ denote the elementwise squares of $v$ and $w$. Both sides of $(\*)$ are nonnegative by the Cauchy-Schwarz inequality, and the LHS gives a non-zero bound for the right RHS, in general.


What strikes me is that the RHS and the LHS of $(\*)$ have different (linear) symmetry groups: the RHS does not change if we apply any orthogonal matrix $U \in \mathbb{R}^{n \times n}$ to both $v$ and $w$, while this is not true for the LHS. Hence, we can strengthen $(\*)$ to
\begin{align\*}
 \sup\_{U^\*U = I}\Big(\|(Uv)^2\| \, \|(Uw)^2\| - \langle (Uv)^2, (Uw)^2 \rangle\Big) \le \|v\|^2 \|w\|^2 - \langle v,w \rangle^2. \quad (\*\*)
\end{align\*}
Unfortunately, I have no idea how to evaluate the LHS of $(\*\*)$.


**Question.** Can we explicitely evaluate the LHS of $(\*\*)$? Or, more generally, is there a version of $(\*)$ for which both sides are invariant under multiplying both $v$ and $w$ by (identical) orthogonal matrices?


Admittedly, this question is a bit vague since it might depend on one's perspective which expressions one considers to be "explicit" and which inequalities one considers to be a "version" of $(\*)$. Nevertheless, I'm wondering whether some people share my intuition that there should be a more symmetric version of $(\*)$.


**Edit.** Maybe it is worthwhile to add the following motivating example: If we choose $n = 2$ and $v = (1,1)/\sqrt{2}$, $w = (1,-1)/\sqrt{2}$, then those vectors are orthogonal and the RHS of $(\*)$ equals $1$, while the LHS of $(\*)$ vanishes since $v^2$ and $w^2$ are linearly dependent. 
However, the LHS of $(\*\*)$ is also equal to $1$; to see this, choose 
$U = \frac{1}{\sqrt{2}}
\begin{pmatrix}
 1 & -1 \\ 1 & 1
\end{pmatrix}
$
and observe that $Uv = (0,1)$, $Uw = (1,0)$.



Symmetric strengthening of the Cauchy-Schwarz inequality
$(\*\*)$ is always equality. By homogenuity we may suppose that $\|v\|=\|w\|=1$, then $Uv,Uw$ are any two unit vectors with prescribed inner product $\langle Uv,Uw\rangle =\langle u,v\rangle$. I claim that you may find two two-dimensional vectors with prescribed given inner product such that equality in $(\*)$ holds. Then you may make $n$-dimensional examples by adding zero coordinates. Well, just look at [GH from MO's proof of (\*)](https://mathoverflow.net/a/301855/53059) and look what we need for equality. If $n=2$, it is sufficient that $F(v,w):=v\_1w\_1-v\_2w\_2=0$. Since $F((v\_1,v\_2),(w\_1,w\_2))=-F((w\_2,w\_1),(v\_2,v\_1))$, we may rotate the vector $(v\_1,v\_2)$ until it becomes equal $(w\_2,w\_1)$, at this moment the vector $(w\_1,w\_2)$ becomes equal to $(v\_2,v\_1)$ (since the inner product and orientation define the pair of unit vectors upto rotation). By continuity $F$ attains zero value at some intermediate point.


