Proof of An Inequality on $\Bbb S^n$ 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.
 A: More generally, let  $x=(x_1,\cdots,x_n)$, $y= (y_1,\cdots,y_n)$ (with $n\geq 2$) such that $\|x\|,\|y\|\leq 1$. Then
$$(x_1y_2-x_2y_1)^2\leq 2(1-\left<x,y\right>).$$
Note that by Cauchy-Schwarz inequality,
$$\begin{align}
(x_1y_2-x_2y_1)^2&=((x_1-y_1)y_2+(y_2-x_2)y_1)^2\\
&\leq \left((x_1-y_1)^2+(x_2-y_2)^2\right)(y_1^2+y_2^2)\\
&\leq \sum_{k=1}^n(x_k-y_k)^2\cdot \|y\|^2\\
&\leq \sum_{k=1}^n(x_k-y_k)^2.
\end{align}$$
On the other hand,
$$2(1-\left<x,y\right>)\geq\sum_{k=1}^nx_k^2+\sum_{k=1}^ny_k^2-2\sum_{k=1}^nx_ky_k=\sum_{k=1}^n(x_k-y_k)^2.$$
A: 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 $\|x-y\|^2 = \|x\|^2-2\langle x, y\rangle +\|y\|^2  = 2(1-\langle x, y\rangle)$.  
