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](https://mathoverflow.net/questions/373553/minimising-the-squared-sum-minus-the-sum-of-squares/373604#373604)



Generalisation of the squared sum minus the sum of squares inequality
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](https://mathoverflow.net/questions/373553/minimising-the-squared-sum-minus-the-sum-of-squares/373604#373604).




---


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.


