Difference between revisions of "1983 USAMO Problems/Problem 2"
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Making such an inequality for all the variable pairs and summing them, we find the lemma is true. | Making such an inequality for all the variable pairs and summing them, we find the lemma is true. | ||
− | Now, | + | Now, let our roots be <math>x_1,x_2,\cdots,x_5</math>. By Vieta's, <math>a=x_1+x_2+\cdots+x_5</math> and <math>b=x_1x_2+x_1x_3+\cdots+x_4x_5</math> |
If we show that for all real <math>x_1,x_2,\cdots, x_5</math> that <math>2a^2\ge 5b</math>, then we have a contradiction and all of <math>x_1,x_2,\cdots, x_5</math> cannot be real. We start by rewriting <math>2a^2\ge 5b</math> as | If we show that for all real <math>x_1,x_2,\cdots, x_5</math> that <math>2a^2\ge 5b</math>, then we have a contradiction and all of <math>x_1,x_2,\cdots, x_5</math> cannot be real. We start by rewriting <math>2a^2\ge 5b</math> as |
Revision as of 17:57, 14 June 2012
1983 USAMO Problem 2
Prove that the zeros of
cannot all be real if .
Solution
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 out the second symmetric sums, and then multiply by to find
which is true by our lemma.