I'd like to know how to show $$\min\_{\Vert x\Vert\_2=1=\Vert y\Vert\_2}\left(\sum\_{k=1}^nx\_ky\_k\right)^2-\sum\_{k=1}^nx\_k^2y\_k^2\geq -1/2.$$


The inequality is discussed in a previous post [Minimum of squared sum minus sum of squares](https://mathoverflow.net/questions/206615/minimum-of-squared-sum-minus-sum-of-squares) but it's not clear to me how to prove this.



Minimising the squared sum minus the sum of squares
It should oftentimes be the case that, analyzing a "thoughtless" Lagrange multiplier solution, one finds a more elegant, "clever" solution. At least, this is the case here. Analyzing the previous Lagrange multiplier solution, one can obtain the following.


We need to show that
$$\sum x\_j^2 y\_j^2\le1/2+\Big(\sum x\_j y\_j\Big)^2\tag{0}$$
given that $\sum x\_j^2=1$ and $\sum y\_j^2=1$.


Let
$$s\_+:=\sum\_{j\colon\,x\_jy\_j>0}x\_jy\_j,\quad s\_-:=-\sum\_{j\colon\,x\_jy\_j<0}x\_jy\_j.$$
Then $s\_+-s\_-=\sum x\_jy\_j$ and $s\_++s\_-=\sum|x\_jy\_j|\le\sqrt{\sum x\_j^2}\sqrt{\sum y\_j^2}=1$, by the Cauchy--Schwarz inequality. Also, $\sum x\_j^2 y\_j^2\le s\_+^2+s\_-^2$.
So, (0) reduces to $s\_+^2+s\_-^2\le1/2+(s\_+-s\_-)^2$, which simplifies to $s\_+ s\_-\le1/4$, which latter holds because $s\_\pm\ge0$ and $s\_++s\_-\le1$. So, (0) is proved.


