# Putnam problem with inequalities

1. Nov 9, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
can someone help me with only if part of Problem 2?

http://www.math.cornell.edu/~putnam/ineqs.pdf

I can prove that quantity is less than or equal to n, but I cannot get it leq to 1?

2. Relevant equations

3. The attempt at a solution

2. Nov 9, 2007

### D H

Staff Emeritus
Assume the first expression is true for all $x_1, x_2, \cdots, x_n$. This means you can hand-pick a particular $x_1, x_2, \cdots, x_n$ that makes the desired relation fall out.

3. Nov 9, 2007

### ehrenfest

Yes. That's how I got that it was less than n. I just picked basis vectors. But I don't know which special vectors pick to get it less than 1.

4. Nov 9, 2007

### D H

Staff Emeritus
You want to leave expressions involving {ri} on both sides of the inequality.

5. Nov 9, 2007

### ehrenfest

So, should I let x equal r times a scalar factor or something? I need to use the Cauchy-Schwartz inequality, somehow, don't I?

6. Nov 9, 2007

### D H

Staff Emeritus
You don't need Cauchy-Schwartz. You are on the right track. Set each xi to ri.

7. Nov 9, 2007

### ehrenfest

So then we have that norm(r) => norm(r)^2 i.e. that 1 => norm(r). I cannot believe how long I spent on that :(

8. Nov 9, 2007

### D H

Staff Emeritus
Almost. You get norm(r)^2 >= norm(r)^4, and thus 1 >= norm(r)^2.

To be completely rigorous you also have to handle the special case where the division in the final step is not valid. This case is trivial, so to speak.

9. Nov 9, 2007

### ehrenfest

When r is a null vector its length is certainly less than or equal to 1.

10. Nov 9, 2007

### D H

Staff Emeritus
Its trivial to deal with the zero vector, but you deal with it you must as step from norm(r)^2 >= norm(4)^r to 1>= norm(r)^2 is not valid for r=0.

11. Nov 9, 2007

### ehrenfest

I agree. Thanks.