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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.

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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, $$2(1-\left<x,y\right>)\geq\sum_{k=1}^nx_k^2+\sum_{k=1}^ny_k^2-2\sum_{k=1}^nx_ky_k=\sum_{k=1}^n(x_k-y_k)^2.$$

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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 $\|x-y\|^2 = \|x\|^2-2\langle x, y\rangle +\|y\|^2 = 2(1-\langle x, y\rangle)$.

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