Question:
The fact that $a^2 \geq 0$ $ \forall a \in \mathbb{R}$; elementary as it may seem, isnevertheless the fundamental idea upon which most important inequalitiesare ultimately based. The great-granddaddy of all inequalities is theSchwarz inequality: $ x_1 y_1 + x_2 y_2 \leq \sqrt {x_1^2 + x_2^2} $ $\sqrt {y_1^2 + y_2^2} $
- Prove that if $x_1 = \lambda y_1 $ and $x_2 = \lambda y_2$ for some number $\lambda$ then equality holds in the Schwarz inequality.
Easy you get $ \lambda (y_1^2 + y_2^2) \geq |\lambda | (y_1^2 + y_2^2) $if we define $ y_1 \geq x_1 $ and $ y_2 >x_2 $ w.o loss of generality both side are equal.
- Prove the same thing when $ y_1 = y_2 = 0$
you just get 0=0 which is fine.
- Now assume that $y_1 $ and $y_2$ are not both $0$ and that there is no such $\lambda $ such that $x_1 = \lambda y_1 $ and $x_2 = \lambda y_2$
Then $ 0 < ( \lambda y_1- x_1)^2 + ( \lambda y_2- x_2)^2 $
How to finish the answer to this part is my question and honestly I have no idea what that last line says/implies and intuitively it looks like gibberish so please dumb down your answer please!
Edit: ( sorry about getting the sign backwards really tired when i wrote this out.) I expanded it $ 0 < \lambda^2 (y_1^2 + y_2^2) -2\lambda ( x_1 y_1 + x_2 y_2) + (x_1^2 + x_2^2)$ not sure if that helps anyone.