Let $x,y$ be two vectors with $\lVert x \rVert = \lVert y \rVert =1$ and $\lVert x-y \rVert \geq \delta$, where $\delta \gt 0$. Is it possible to show that, $1-(x^Ty)^2 \geq \delta^2$?
My Approach:
$$\begin{align}\lVert x-y \rVert \geq \delta &\implies \lVert x-y \rVert^2 \geq \delta^2 \\&\implies (x-y)^T(x-y) \geq \delta^2 \\&\implies (\lVert x \rVert^2+\lVert y \rVert^2-2x^Ty) \geq \delta^2 \\&\implies 2(1-x^Ty) \geq \delta^2\end{align}$$
I also tried to apply the Cauchy-Schwartz inequality and triangle inequality, but couldn't reach the final statement.
