TITLE: Proof of An Inequality on $\Bbb S^n$
QUESTION [5 upvotes]: I have no idea for proving the following inequality:

Let $x=(x_1,\cdots,x_n), y= (y_1,\cdots,y_n)\in\Bbb S^n$. then 
  $$(x_1y_2-x_2y_1)^2\leq 2(1-\left<x,y\right>).$$

Any help would be great.
Thanks.

REPLY [4 votes]: Consider the matrix
\begin{align}
A = 
\begin{pmatrix}
0 & -1 & 0 &\ldots & 0\\
1 & 0 & 0 &\ldots & 0\\
0 & 0 & 0 &\dots & 0\\
\vdots & \vdots &\vdots &\ddots &\vdots\\
0 & 0& 0 &\ldots & 0
\end{pmatrix}
\end{align}
and observe that
\begin{align}
y^TA x =  x_1y_2 - x_2y_1 \ \ \text{ and } \ \  \ x^TAx = y^TAy=0.
\end{align}
Next, observe that
\begin{align}
y^TAx = y^TA(x-y)
\end{align}
then by Cauchy-Schwarz inequality we get that
\begin{align}
|y^TAx|^2 \leq \|A\|^2\|y\|^2\|x-y\|^2 = \|A\|^2\|x-y\|^2. 
\end{align}
Lastly, it's not hard to see that $\|A\| = 1$ and $\|x-y\|^2 = \|x\|^2-2\langle x, y\rangle +\|y\|^2  = 2(1-\langle x, y\rangle)$.