I'm really bad at analysis and this problem was recommend to me to help me grasp some basics of $\epsilon $ $\delta $ So I'm doing a problem (though it's like 12 pieces) this is I guess the fourth part.

Prove 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} $)

using $ 2xy \leq x^2 + y^2 $ and

$ x= \dfrac {x_i} { \sqrt {x_1^2 + x_2^2}}$, $ y= \dfrac {y_i} { \sqrt {y_1^2 + y_2^2}}$

First for $i=1$ and then for $i=2$

I'm not really following what there saying here so I just subed in the two cases but that gave me two different expressions. Since I knew what I wanted to get I added the two case together and it happened to reduce to the Schwartz inequality.

$i=1$ case $ \dfrac {2x_1 y_1} {\sqrt {y_1^2 + y_2^2} \sqrt {x_1^2 + x_2^2}} \leq \dfrac {x_1^2} {x_1^2 + x_2^2} + \dfrac {y_1^2} {y_1^2 + y_2^2}$

$i=2$ case $ \dfrac {2x_2 y_2} {\sqrt {y_1^2 + y_2^2} \sqrt {x_1^2 + x_2^2}} \leq \dfrac {x_2^2} {x_1^2 + x_2^2} + \dfrac {y_2^2} {y_1^2 + y_2^2}$

adding both cases together we get

$ 2(\dfrac {x_1 y_1 + x_2 y_2} {\sqrt {x_1^2 + x_2^2} \sqrt {y_1^2 + y_2^2}}) \leq \dfrac {x_1^2 + x_2^2} {x_1^2 + x_2^2} + \dfrac {y_1^2 + y_2^2} {y_1^2 + y_2^2} $

Edit: I added this step cause someone thinks $ x_1 =x_2 =0 $ and $ y_1 =y_2 =0 $ needs to be taken into account but this shows that each of them =1 $ \forall x_1, x_2, y_1, y_2 \in \mathbb{R} $

$ (\dfrac {x_1 y_1 + x_2 y_2} {\sqrt {x_1^2 + x_2^2} \sqrt {y_1^2 + y_2^2}}) \leq 1 $

thus we have $ x_1 y_1 + x_2 y_2 \leq \sqrt {x_1^2 + x_2^2} $ $\sqrt {y_1^2 + y_2^2} $

Now I just need to show that $ 2xy \leq x^2 + y^2 $ is actually true since its equivalent to say $ 0 \leq x^2 + y^2 -2xy$ or $0 \leq ( x-y)^2$ this becomes obvious the trivial case x=y=0 is the only time we have equality and for all other values of x and y $(x-y)^2 > 0$

I guess that concludes the proof but I still have some questions.

The first one being is this proof rigorous enough?

Secondly I'm having some trouble with what exactly/ where exactly this inequality is coming from/means. What would lead someone to that guess for the values for $x_i$?

Lastly I am having trouble with what $(x-y)^2 $ looks like I mean I can picture $ 0 \leq x^2 +y^2 <5^2 $ ( its a circular surface ) when i try and picture $5^2 >(x-y)^2 \geq 0 $ I feel like I'm looking at a surface trapped between two bounds on a funny angle is that what it looks like? ( look at the picture) The picture

  • $\begingroup$ It's rigorous except you can't divide by zero, so you need to handle those special cases separately. Also, you didn't even state what $x_i$ and $y_i$ are, I assume they are real. And why do you talk about the equality in $0\le(x-y)^2$? $x=y=0$ isn't the only case, there's quite a lot other cases. Are you supposed to find when the equality is attained in the CS inequality? Why do you need to visualize the $(x-y)^2$ expression? The only thing you need is that it's not negative, the shape is irrelevant. $\endgroup$ – user2345215 Dec 19 '14 at 18:50
  • $\begingroup$ i thought i already excluded the case when x=y=0 there is no other time that's the right side =0 is there? me picturing it is just me trying to understand whats actually going on. i feel like this inequality is some mythical creature if i can't understand what it means ill never be able to apply it. $\endgroup$ – Faust Dec 19 '14 at 18:57
  • $\begingroup$ And i Specifically asked WHAT $ x_i $ and $ y_i$ are as one of my questions my textbook just rights them down exactly as i did in the question. $\endgroup$ – Faust Dec 19 '14 at 18:58
  • $\begingroup$ I don't understand you at all. Maybe someone else will help you. $\endgroup$ – user2345215 Dec 19 '14 at 19:01
  • $\begingroup$ i added anther step to show why i think that x=y=0 is the only case i need to exempt. $\endgroup$ – Faust Dec 19 '14 at 19:09

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