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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.

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    $\begingroup$ The direction should be reversed $x_1 y_1 + x_2 y_2 \le \sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}$ $\endgroup$ – r9m Dec 19 '14 at 3:58
  • $\begingroup$ $0 < ( \lambda y_1- x_1)^2 + ( \lambda y_2- x_2)^2$ is a quadratic in $\lambda$ and positive ! What's the discriminant of this quadratic ? :) $\endgroup$ – r9m Dec 19 '14 at 4:03
  • $\begingroup$ "the great-granddaddy of all inequalities" is the triangle inequality imho $\endgroup$ – GFauxPas Dec 19 '14 at 4:13
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    $\begingroup$ @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) $\endgroup$ – Matt Samuel Dec 19 '14 at 4:19
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    $\begingroup$ @r9m 11$ from Amazon hell yeah i bought it =) $\endgroup$ – Faust Dec 19 '14 at 5:03
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As you obtained,$$0 < ( \lambda y_1- x_1)^2 + ( \lambda y_2- x_2)^2 \iff (y_1^2+y_2)^2\lambda^2-2(x_1y_1+x_2y_2)\lambda+(x_1^2+x_2^2) > 0$$

Now if you are not familiar with using the discriminant, you can "complete the square" to get: $$\iff (y_1^2+y_2^2)\left(\color{red}{\lambda - \frac{x_1y_1+x_2y_2}{y_1^2+y_2^2}} \right)^2+\color{blue}{(x_1^2+x_2^2)-\frac{(x_1y_1+x_2y_2)^2}{(y_1^2+y_2^2)}} > 0$$

As this must be true for all $\lambda$, in particular it must hold for $\color{red}{\lambda = \dfrac{x_1y_1+x_2y_2}{y_1^2+y_2^2}} $, from which the desired inequality follows. $$\implies \color{blue}{(x_1^2+x_2^2)-\frac{(x_1y_1+x_2y_2)^2}{(y_1^2+y_2^2)}} > 0 \implies (x_1^2+x_2^2)(y_1^2+y_2^2)> (x_1y_1+x_2y_2)^2$$

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