In this great question 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 $Uv = (0,1)$, $Uw = (1,0)$.