Difference between revisions of "1983 USAMO Problems/Problem 2"
Danielguo94 (talk | contribs) (→Solution) |
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By the trivial inequality, | By the trivial inequality, | ||
− | <cmath>x^2+y^2\ge 2xy \ | + | <cmath>x^2+y^2\ge 2xy \Rightarrow \frac{x^2}{2} + \frac{y^2}{2} \ge xy</cmath> |
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, we start by plugging in our Vieta's: Let our roots be | + | Now, we start by plugging in our Vieta's: Let our roots be <math>x_1,x_2,\cdots,x_5</math>. This means be Vieta's that <math>a=x_1+x_2+\cdots+x_5, 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 17:55, 14 June 2012
1983 USAMO Problem 2
Prove that the zeros of
cannot all be real if .
Solution
Lemma:
For all real numbers ,
We solve this cyclically by showing
By the trivial inequality,
Making such an inequality for all the variable pairs and summing them, we find the lemma is true.
Now, we start by plugging in our Vieta's: Let our roots be . This means be Vieta's that
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.