Symmetric strengthening of the Cauchy-Schwarz inequality

Question

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 $Uv = (0,1)$, $Uw = (1,0)$.

Answer 1

$(**)$ 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 (*) 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.

Answer 2

This is not an answer, rather a heuristic argument.

I have done some numerical experiments for pairs of randomly chosen $v,w$, where I have used products of random Householder matrices for the $U$'s. They suggest that the LHS can come arbitrarily close to the RHS. So there should be a set of extremal unitary matrices $U$ for which equality is attained. For $n=2$ this $U$ seems unique up to sign (but the equation for it is already rather messy for the general case of $v,w$), and for $n\geqslant3$ there seems to be a whole bunch of them. (a continuum? or rather a finite number of isolated matrices?)

By homogeneity, we can assume wlog $\|v\|= \|w\|=1$. Then it would be an interesting question to what those extremal unitary matrices, which are defined only by the pair $\{v,w\}\in\mathbb R^n\times\mathbb R^n$, correspond geometrically.