1983 USAMO Problems/Problem 2

Problem

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

We prove the contrapositive: if the polynomial in question has the five real roots $x_1, x_2, x_3, x_4, x_5$ , then $5b \le 2a^2$ .

Because $a = -(x_1 + x_2 + x_3 + x_4 + x_5)$ and $b = x_1x_2 + x_1x_3 + ... + x_4x_5$ by Vieta's Formulae, we have

$2b = 2x_1x_2 + 2x_1x_3 + ... + 2x_4x_5 = (x_1 + x_2 + x_3 + x_4 + x_5)^2 - (x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2)$ $=a^2 - \frac{(1+1+1+1+1)(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2)}{5}$ $\le a^2 - \frac{(x_1 + x_2 + x_3 + x_4 + x_5)^2}{5}$ $=\frac{4a^2}{5},$

so $5b \le 2a^2$ , as desired.

Solution 2

Lemma:

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

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

$x_1x_2+x_1x_3+\cdots+x_4x_5$

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, let our roots be $x_1,x_2,\cdots,x_5$ . By Vieta's, $a=x_1+x_2+\cdots+x_5$ and $b=x_1x_2+x_1x_3+\cdots+x_4x_5$

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 the sum in brackets, 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.

1983 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5
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1983 USAMO Problems/Problem 2

Contents

Problem

Solution

Solution 2

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