# The equality case of the Schwartz inequality

Question:

The fact that $a^2 \geq 0$ $\forall a \in \mathbb{R}$; elementary as it may seem, is nevertheless the fundamental idea upon which most important inequalities are ultimately based. The great-granddaddy of all inequalities is the Schwarz inequality: $x_1 y_1 + x_2 y_2 \leq \sqrt {x_1^2 + x_2^2}$ $\sqrt {y_1^2 + y_2^2}$

1. 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.

1. Prove the same thing when $y_1 = y_2 = 0$

you just get 0=0 which is fine.

1. 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.

• The direction should be reversed $x_1 y_1 + x_2 y_2 \le \sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}$ – r9m Dec 19 '14 at 3:58
• $0 < ( \lambda y_1- x_1)^2 + ( \lambda y_2- x_2)^2$ is a quadratic in $\lambda$ and positive ! What's the discriminant of this quadratic ? :) – r9m Dec 19 '14 at 4:03
• "the great-granddaddy of all inequalities" is the triangle inequality imho – GFauxPas Dec 19 '14 at 4:13
• @GFauxPas Apparently it was more than one author, and foolish and young as I was at the time I didn't realize how famous they were. by G. H. Hardy (Author), J. E. Littlewood (Author), G. Pólya (Author) – Matt Samuel Dec 19 '14 at 4:19