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, 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> |
| − | |||
| − | <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 18:16, 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.
