# 1983 USAMO Problems/Problem 2

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

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.

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