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Difference between revisions of "1983 USAMO Problems/Problem 2"
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Now, we start by plugging in our Vieta's: Let our roots be  
 
Now, we start by plugging in our Vieta's: Let our roots be  
  
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<math>x_1,x_2,\cdots,x_5</math>. This means be Vieta's that
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<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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<math>a=x_1+x_2+\cdots+x_5, 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 18:16, 13 November 2011

1983 USAMO Problem 1

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

\[x_1^2+x_2^2+\cdots+x_5^2+2(x_1x_2+x_1x_3+\cdots+x_4x_5)\ge\frac{5}{2}(x_1x_2+x_1x_3+\cdots+x_4x_5)\]

Aha! 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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