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

Revision as of 18:28, 13 November 2011

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 $x_1,x_2,\cdots, x_3$ ,

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

$x_1x_2+x_1x_3+\cdots+x_4x_5$

We solve this cylicallly by showing

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

By the trivial inequality, $(x-y)^2\ge 0$ , or $x^2+y^2-2xy\ge 0$ .

$x^2+y^2\ge 2xy$

Dividing by $2$ 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 $x_1,x_2,\cdots,x_5$ . 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 43: / Line 43: @@
 <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 by <math>2</math> to find
+We subtract out the second symmetric sums, and then multiply by <math>2</math> to find
 <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.

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

Revision as of 18:28, 13 November 2011

1983 USAMO Problem 2

Solution