TITLE: When does Schwarz inequality become an equality?
QUESTION [10 upvotes]: In Spivak Calculus you are asked to prove that in Schwarz inequality, equality holds only when $y_1 = y_2 = 0$ or when there is a number $\lambda$ such that $x_1 = \lambda y_1$ and $x_2 = \lambda y_2$.
I can go from $x_1y_1 + x_2y_2 = \sqrt{x_1^2 + x_2^2} + \sqrt{y_1^2 + y_2^2}$ to $x_1y_2 = x_2y_1$ but then I got stuck, or is it that the implication is the other way around?

REPLY [2 votes]: Since
$$
\left(x_1^2+x_2^2\right)\left(y_1^2+y_2^2\right)-(x_1y_1+x_2y_2)^2=(x_1y_2-x_2y_1)^2
$$
we get equality if and only if $x_1y_2=x_2y_1$.