# Putnam problem with inequalities

## 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?

## The Attempt at a Solution

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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.

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

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

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
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.

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

D H
Staff Emeritus