Putnam problem with inequalities

[ehrenfest]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

 

[D H]

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.

 

[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.

 

[D H]

You want to leave expressions involving {r_i} on both sides of the inequality.

 

[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?

 

[D H]

You don't need Cauchy-Schwartz. You are on the right track. Set each x_i to r_i.

 

[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 :(

 

[D H]

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.

 

[ehrenfest]

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

 

[D H]

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.

 

[ehrenfest]

I agree. Thanks.