Question

PROVE THAT magnitude(u+v) 11 years ago

Answer 1

This is the Cauchy Schwarz Inequality.
\[\begin{align*}|u+v|^2&=(u+v)\cdot(u+v)\\
&=u\cdot u+v\cdot u+u\cdot v+v\cdot v\\
&=|u|^2+2(u\cdot v)+|v|^2\\
&\le |u|^2+2|u||v|+|v|^2\\
&\le(|u|+|v|)^2\\
\sqrt{|u+v|^2}&\le\sqrt{(|u|+|v|)^2}\\
|u+v|&\le|u|+|v|\end{align*}\]