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a difficult inequality
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skywalkerJ.L.
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#1
a difficult inequality

Prove that for n distinct positive integers a_1,a_2,\ldots,a_n,the following inequality holds:
{a_1}^3+{a_2}^3+...{a_n}^3\ge (a_1+a_2+\ldots+a_n)^2

PostPosted: Thu Nov 05, 2009 8:26 am  Back to top 
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leshik
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#2
Simple proof can be based on induction. To prove the step from n-1 to n it is enough to show that a^2_n\ge a_n +2(\sum\limits_{i=1}^{n-1} a_i). Now just use obvious inequalities a_{n-i}\le a_n -i and a_n\ge n Wink

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MeKnowsNothing
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#3
Am I right that because of Newton's sums, one can reduce this to three variable inequality?

PostPosted: Thu Nov 05, 2009 10:38 am  Back to top 
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skywalkerJ.L.
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#4
MeKnowsNothing wrote:
Am I right that because of Newton's sums, one can reduce this to three variable inequality?


Newton's sum?
What is the content of this theorem?

By the way,thanks for leshik, i finally prove it because of your hint.

PostPosted: Sat Nov 07, 2009 2:15 am  Back to top 
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MeKnowsNothing
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#5
skywalkerJ.L. wrote:
MeKnowsNothing wrote:
Am I right that because of Newton's sums, one can reduce this to three variable inequality?


Newton's sum?
What is the content of this theorem?

By the way,thanks for leshik, i finally prove it because of your hint.

My idea was to consider to polynomial (x - a_1)\cdot\ldots\cdot (x - a_n). Then sum of a_i's and a_i^3's can be represented as a rational expressions with three unknowns.

PostPosted: Sat Nov 07, 2009 5:33 am  Back to top 
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