1983 USAMO Problems/Problem 2
Contents
Problem
Prove that the zeros of
cannot all be real if .
Solution
We prove the contrapositive: if the polynomial in question has the five real roots , then .
Because and by Vieta's Formulae, we have
(by Cauchy-Schwarz)
so , as desired.
Solution 2
Lemma:
For all real numbers ,
By the trivial inequality,
Making such an inequality for all the variable pairs and summing them, we find the lemma is true.
Now, let our roots be . By Vieta's, and
If we show that for all real that , then we have a contradiction and all of cannot be real. We start by rewriting as
We divide by and find
Expanding the LHS, we have
We subtract the sum in brackets, and then multiply by to find
which is true by our lemma.
See Also
1983 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.