Generalisation of the squared sum minus the sum of squares inequality I'd like to know how to prove
$$\min_{\Vert x\Vert_2=1=\Vert y\Vert_2}\left|\sum_{k=1}^nx^*_ky_k\right|^2-\sum_{k=1}^n|x_k|^2|y_k|^2\geq -1/2$$
for $x,y\in\mathbb{C}^n$ with $\Vert x\Vert_2=\Vert y\Vert_2=1$.
This is a generalisation from $\mathbb{R}^n$ to $\mathbb{C}^n$ of the inequality which was proved in Minimising the squared sum minus the sum of squares
 A: Replacing $x_k^*$ by $x_k$ and noting that $|x_k^*|=|x_k|$, we see that the problem is to show that
$$|s|^2-\sum_{k=1}^n|x_k|^2|y_k|^2\ge-1/2$$
for $x,y\in\mathbb{C}^n$ with $\|x\|_2=\|y\|_2=1$, where
$$s:=\sum_{k=1}^n v_k,\quad v_k:=x_ky_k.$$
For any fixed values of the $|x_j|$'s and $|y_k|$'s (so that values of the $|v_k|$'s are also fixed), minimize $|s|^2$. Let $((x_j),(y_k))$ be a corresponding minimizer. Note that
$$|s|^2\ge(|v_1|-|s-v_1|)^2=(|v_1|-|v_2+\cdots+v_n|)^2,$$
with the equality iff either (i) the 2D vectors $v_1$ and $s-v_1$ are both nonzero and have the opposite directions or (ii) at least one of them is the zero vector. So, by also fixing the vectors $v_2,\dots,v_n$, we see that, for any minimizing $((x_j),(y_k))$, the vector $v_1$ must be collinear with the vector $s$. Similarly, all the $v_j$'s must be collinear with the vector $s$.
So, if $s\ne0$, then all the vectors $v_k:=x_ky_k$ are collinear, and the problem reduces to the case when the $x_j$'s and $y_k$'s are real numbers, and in this case the problem was solved.

The remaining problem, corresponding to the case $s=0$, is this: show that
$$\sum_{k=1}^n|x_k|^2|y_k|^2\le1/2$$
for $x,y\in\mathbb{C}^n$ with $\|x\|_2=\|y\|_2=1$ and $\sum_{k=1}^n x_ky_k=0$.
The condition $\sum_{k=1}^n x_ky_k=0$ obviously implies $|x_jy_j|\le\sum_{k\in[n]\setminus\{j\}}|x_ky_k|$ -- or, equivalently, $2|x_jy_j|\le\sum_{k\in[n]}|x_ky_k|$ -- for all $j\in[n]:=\{1,\dots,n\}$.
So, letting $a_j:=|x_j|$ and $b_j:=|y_j|$, we see that the problem reduces to proving

Claim: If $a_j\ge0$, $b_j\ge0$, and $2a_jb_j\le\sum_{i\in[n]}a_ib_i$ for all $j\in[n]$, and if $\sum_{j\in[n]}a_j^2=1=\sum_{j\in[n]}b_j^2$, then $\sum_{j\in[n]}a_j^2b_j^2\le1/2$.

The proof of this claim is easy. Indeed, by the Cauchy--Schwarz inequality, $\sum_{i\in[n]}a_ib_i\le1$. So, for all $j\in[n]$ we have $2a_jb_j\le\sum_{i\in[n]}a_ib_i\le1$ and hence $a_jb_j\le1/2$ and $\sum_{j\in[n]}a_j^2b_j^2\le\frac12\,\sum_{j\in[n]}a_jb_j\le1/2$.
This completes the proof of the Claim and thus the entire proof of the desired inequality.
