We have the proof of triangle inequality. I need to change the proof in some way to get the Pythagorean theorem : $∥u+v∥ =∥u∥^2 +∥v∥^2$. How to do this?
Triangle inequality:
$∥u+v∥^2 =(u+v)·(u+v)$
= $(u + v) · u + (u + v) · v$
= $u · u + v · u + u · v + v · v $
= $∥u∥^2 +u·v+u·v+∥v∥^2 $
= $∥u∥^2 + 2(u · v) + ∥v∥^2 $
≤ $∥u∥^2 +2|u·v|+∥v∥^2$
≤ $∥u∥^2 + 2 ∥u∥ ∥v∥ + ∥v∥^2$
= $(∥u∥ + ∥v∥)^2$
In the setting of the Pythagorean Theorem, $u$ and $v$ are perpendicular vectors. Therefore, $u\cdot v=v\cdot u=0$ and you can conclude after your fourth equality.