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 (*) 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.

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.