Assume$$x_1y_1 + x_2y_2 \leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}$$
Prove that if $ x_1 = \lambda y_1 $ and $ x_2 = \lambda y_2 $ for some number $ \lambda $ then equality holds in the above statement. Prove the same thing of $ y_1 = y_2 = 0 $.

Question

Assume$$x_1y_1 + x_2y_2 \leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}$$
Prove that if $ x_1 = \lambda y_1 $  and  $ x_2 = \lambda y_2 $  for some number $ \lambda $ then equality holds in the above statement. Prove the same thing of $ y_1 = y_2 = 0 $.

anonymous · Answer

just plug in the values, right?

mr.math · Answer

Yep!

anonymous · Answer

anyway, what's so special about the first statement?

mr.math · Answer

It's trivial.

mr.math · Answer

Nothing.

anonymous · Answer

because later in the question it asks us to prove it!

anonymous · Answer

and I think that lambda and y = 0 part is the first step in proving it :-P

mr.math · Answer

You mean the inequality?

anonymous · Answer

yeah

mr.math · Answer

Hmm, maybe it wants you to consider the different cases of x and y.

mr.math · Answer

The equality holds too for $x_1=x_2=0$. So you're left with only four cases,
Cases 1: $x_1=0$ and other different from zero.
Case 2: $x_2=0$ \ \ \ .
You've got the idea I think.

jamesj · Answer

It's an example of an extremely important inequality, probably the most important in mathematics: the Cauchy-Schwarz inequality.

This statement you have of it says that if x = (x1,x2) and y = (y1,y2) are vectors in R^2, then their inner product is less than or equal to their (Euclidean) norm,
$$ x . y \ \leq \ ||x||_2 \  ||y||_2 $$

anonymous · Answer

It's a very early question in the Spivak book, and it already touches on this idea :-D

anonymous · Answer

but it didn't really explain it's significance in the question :(

anonymous · Answer

it only asks us to prove that it is true using what we have already learned

jamesj · Answer

It's warming you up so when he does get to it you've seen it already.

jamesj · Answer

and on the contrary, Spivak called it the Schwarz inequality and even points you towards a more general form (Problem 2-21).  So really, no complaining if you didn't bother to read that!

anonymous · Answer

oh :( but I should work on the remaining questions before I head on to chapter 2 :-D

jamesj · Answer

wouldn't do you any harm.