1983 USAMO Problems/Problem 2


Prove that the zeros of


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


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


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



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


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.

See Also

1983 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5
All USAMO Problems and Solutions

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