Difference between revisions of "1983 USAMO Problems/Problem 2"

Revision as of 17:55, 14 June 2012

1983 USAMO Problem 2

Prove that the zeros of

$x^5+ax^4+bx^3+cx^2+dx+e=0$

cannot all be real if $2a^2<5b$ .

Solution

Lemma:

For all real numbers $x_1,x_2,\cdots x_5$ ,

$2(x_1^2+x_2^2+\cdots+x_5^2)\ge$

$x_1x_2+x_1x_3+\cdots+x_4x_5$

We solve this cyclically by showing

$\frac{1}{2}x^2+\frac{1}{2}y^2\ge xy$

By the trivial inequality,

$x^2+y^2\ge 2xy \Rightarrow \frac{x^2}{2} + \frac{y^2}{2} \ge xy$

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 $x_1,x_2,\cdots,x_5$ . This means be Vieta's that $a=x_1+x_2+\cdots+x_5, b=x_1x_2+x_1x_3+\cdots+x_4x_5$

If we show that for all real $x_1,x_2,\cdots, x_5$ that $2a^2\ge 5b$ , then we have a contradiction and all of $x_1,x_2,\cdots, x_5$ cannot be real. We start by rewriting $2a^2\ge 5b$ as

$2(x_1+x_2+\cdots+x_5)^2\ge 5(x_1x_2+x_1x_3+\cdots+x_4x_5)$

We divide by $2$ and find

$(x_1+x_2+\cdots+x_5)^2\ge \frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)$

Expanding the LHS, we have

$x_1^2+x_2^2+\cdots+x_5^2+2(x_1x_2+x_1x_3+\cdots+x_4x_5)\ge\frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)$

We subtract out the second symmetric sums, and then multiply by $2$ to find

$2x_1^2+2x_2^2+\cdots+2x_5^2\ge x_1x_2+x_1x_3+\cdots+x_4x_5$

which is true by our lemma.

@@ Line 23: / Line 23: @@
 By the trivial inequality,
-<cmath>x^2+y^2\ge 2xy \rightarrow \frac{x^2}{2} + \frac{y^2}{2} \ge xy</cmath>
+<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.
-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>
+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

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Difference between revisions of "1983 USAMO Problems/Problem 2"

Revision as of 17:55, 14 June 2012

1983 USAMO Problem 2

Solution