In order to show that , one assumes without loss of generality . By expanding

$\displaystyle \|x-t\alpha y\|^2\geq 0$

where and , one can not only conclude that the Cauchy inequality is true *but also* that the equality is true only when are linearly dependent.

The choice of and seem rather mysterious; one the other hand, the way one chooses and in this excellent post is extraordinarily natural. However, maybe I miss something in the post; I don’t see how to conclude the “equality” part from your argument.

]]>