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Difference between revisions of "1983 USAMO Problems/Problem 2"
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Making such an inequality for all the variable pairs and summing them, we find the lemma is true.
 
Making such an inequality for all the variable pairs and summing them, we find the lemma is true.
  
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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>
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Now, let our roots be <math>x_1,x_2,\cdots,x_5</math>. By Vieta's, <math>a=x_1+x_2+\cdots+x_5</math> and <math>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
 
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

Revision as of 16:57, 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 $x_1,x_2,\cdots x_5$,

\[2(x_1^2+x_2^2+\cdots+x_5^2)\ge\]

\[x_1x_2+x_1x_3+\cdots+x_4x_5\]

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 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.

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