Difference between revisions of "1983 USAMO Problems/Problem 2"
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Dividing by <math>2</math> gives us the desired. | Dividing by <math>2</math> gives us the desired. | ||
| − | Making such an inequality for all the variable pairs and summing | + | Making such an inequality for all the variable pairs and summing them, we find the lemma is true. |
| − | |||
| − | 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 | ||
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<math>a=x_1+x_2+\cdots+x_5, b=x_1x_2+x_1x_3+\cdots+x_4x_5</math> | <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>, | + | 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 |
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| − | 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 | ||
| − | <cmath>2(x_1+x_2+\cdots+x_5)^2\ge | + | <cmath>2(x_1+x_2+\cdots+x_5)^2\ge 5(x_1x_2+x_1x_3+\cdots+x_4x_5)</cmath> |
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We divide by <math>2</math> and find | We divide by <math>2</math> and find | ||
| − | <cmath>(x_1+x_2+\cdots+x_5)^2\ge | + | <cmath>(x_1+x_2+\cdots+x_5)^2\ge \frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)</cmath> |
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| − | |||
Expanding the LHS, we have | Expanding the LHS, we have | ||
| − | <cmath>x_1^2+x_2^2+\cdots+x_5^2+2(x_1x_2+x_1x_3+\cdots+x_4x_5)\ge | + | <cmath>x_1^2+x_2^2+\cdots+x_5^2+2(x_1x_2+x_1x_3+\cdots+x_4x_5)\ge\frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)</cmath> |
| − | |||
| − | |||
Aha! We subtract out the second symmetric sums, and then multiply | Aha! We subtract out the second symmetric sums, and then multiply | ||
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by <math>2</math> to find | by <math>2</math> to find | ||
| − | <cmath>2x_1^2+2x_2^2+\cdots+2x_5^2\ge | + | <cmath>2x_1^2+2x_2^2+\cdots+2x_5^2\ge x_1x_2+x_1x_3+\cdots+x_4x_5</cmath> |
| − | |||
| − | |||
which is true by our lemma. | which is true by our lemma. | ||
Revision as of 18:15, 13 November 2011
1983 USAMO Problem 1
Prove that the zeros of
![\[x^5+ax^4+bx^3+cx^2+dx+e=0\]](http://latex.artofproblemsolving.com/9/0/4/9049ba78fdefb11bd3a89101dac429cad1017dce.png)
cannot all be real if 
.
Solution
Lemma:
For all real numbers 
,
![\[2(x_1^2+x_2^2+\cdots+x_5^2)\ge\]](http://latex.artofproblemsolving.com/2/a/f/2af6290fdbbe17fe31041fbabb8fdd8af5f57b68.png)
![\[x_1x_2+x_1x_3+\cdots+x_4x_5\]](http://latex.artofproblemsolving.com/2/3/b/23bbfb6c76eb1d1a85d14e0666b3ad01ddbc9321.png)
We solve this cylicallly by showing
![\[\frac{1}{2}x^2+\frac{1}{2}y^2\ge xy\]](http://latex.artofproblemsolving.com/8/3/9/83965288b295c072e1166741020573313dbbbb8b.png)
By the trivial inequality, 
, or 
.
![\[x^2+y^2\ge 2xy\]](http://latex.artofproblemsolving.com/6/c/a/6ca3dd65c224b1def47a4afb2f33f52ede55be21.png)
Dividing by 
gives us the desired.
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
![\[2(x_1+x_2+\cdots+x_5)^2\ge 5(x_1x_2+x_1x_3+\cdots+x_4x_5)\]](http://latex.artofproblemsolving.com/6/8/c/68cd8c64a170cac38a6a379e1dcbeea283fe6af6.png)
We divide by and find
![\[(x_1+x_2+\cdots+x_5)^2\ge \frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)\]](http://latex.artofproblemsolving.com/2/2/4/224cefe651caf9d583b8ae9437edc9f0bed7166f.png)
Expanding the LHS, we have
![\[x_1^2+x_2^2+\cdots+x_5^2+2(x_1x_2+x_1x_3+\cdots+x_4x_5)\ge\frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)\]](http://latex.artofproblemsolving.com/e/0/2/e02af40fa9b7b59545e1d15374ec9ef9ebd3762c.png)
Aha! We subtract out the second symmetric sums, and then multiply
by to find
![\[2x_1^2+2x_2^2+\cdots+2x_5^2\ge x_1x_2+x_1x_3+\cdots+x_4x_5\]](http://latex.artofproblemsolving.com/2/c/0/2c0f1bf151e7b7846d2ec859820ab4461d442cb6.png)
which is true by our lemma.
