# Prove (xu + yv )^2 <= (x^2 + y^2)(u^2 + v^2)

1. Aug 26, 2010

### NastyAccident

1. The problem statement, all variables and given/known data
(!) Let x, y, u, v be real numbers
a.) Prove (xu + yv )^2 <= (x^2 + y^2)(u^2 + v^2)
b.) Determine precisely when equality holds in part (a).

2. Relevant equations
Looks like the cauchy-stewarts inequality:
(a*b)^2 <= a*b

3. The attempt at a solution
0 <= (x^2+y^2)(u^2+v^2) - (xu +yv )^2
0 <= x^2u^2 + x^2v^2 + y^2u^2 + y^2v^2 - (x^2u^2 + 2xuyv + y^2v^2)
0 <= x^2v^2 + y^2u^2 - 2xuyv
0 <= (xv - yu)^2

Any square is always greater than 0.

However, I'm relatively new at trying to do proofs and the book's (Mathematical Thinking Problem Solving and Proofs) examples do not use an already established inequality to prove it. They typically start from scratch. Also, this problem set was pretty much the instructor kicking us from the shallow end of a pool into the deep deep end without any floaties.

In any case, I'm not sure about how to even approach part B if part a holds.

Unless it is, for all real numbers since the square allows for the terms inside to be negative.

Sincerely,

NastyAccident
One confused Mathematics Major.

2. Aug 26, 2010

### hunt_mat

You've aced part a) part b is even simplier it is equal when xv=yu, or that ratios:
$$\frac{x}{y}=\frac{u}{v}$$

3. Aug 26, 2010

### NastyAccident

So,

You pretty much went one step further and....
0 <= (xv - yu)^2
0 <= xv - yu
yu <= xv

Correct?

So, in the future just take it one step further and figure out what the zeros are to figure out when the equality 'holds.'

Thanks for the help by the way!

Sincerely,

NastyAccident

4. Aug 26, 2010

### hunt_mat

Yes.

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