Why is Proving Equality in the Schwarz Inequality Problem Challenging?

[zooxanthellae]

Homework Statement

From Spivak's Calculus Chapter 1:

"Suppose that $$y_1$$ and $$y_2$$ are not both $$0$$, and that there is no number λ such that $$x_1 =$$ λ$$y_1$$ and $$x_2 =$$ λ$$y_2$$."

Then $$0$$<(λ$$y_1 - x_1)^2 + ($$λ$$y_2 - x_2)^2$$.

Using problem 18 (which involved proofs related to inequalities like $$x^2 + xy + y^2$$), complete the proof of the Schwarz Inequality.

Homework Equations

None strike me.

The Attempt at a Solution

The thing that's really bothering me about this is that the problem I've given is just part a) of the problem. In part d) I am asked to "Deduce...that equality holds only when $$y_1 = y_2 = 0$$ or when there is a number λ $$\geq 0$$ such that $$x_1 =$$ λ$$y_1$$ and $$x_2 =$$ λ$$y_2$$. Well, in a) he asked me to assume that both of those things were not true to start my proof. Doesn't this mean that, starting with those conditions, one cannot prove that equality is possible, and thus one can't prove the entirety of the Schwarz inequality (as in, the less than or equal to part)?

 

[HallsofIvy]

Yes assuming what is given, you could prove that [math]|<x, y>|< <x, x><y, y>[/math] while the "Cauchy-Schwarts" inequality only asserts "$\le$".

However, if you could prove "less than" you would have proved "less than or equal two". The latter is a subset of the former.

 

[zooxanthellae]

I guess my issue is that when I think of proving that something is "less than or equal to" something else, I feel like I have to prove that it could be either less than or equal to that something else. It feels...sloppy not to. I guess that's my hang-up, though.

 

[HallsofIvy]

It is certainly correct to say that "3 is less than or equal to 4". Of course, that is not a very "sharp" inequality!