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1991 USAMO Problems/Problem 4 - Revision history
2024-03-29T12:09:00Z
Revision history for this page on the wiki
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1=2: /* Resources */
2016-07-19T11:39:26Z
<p><span dir="auto"><span class="autocomment">Resources</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 11:39, 19 July 2016</td>
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<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>== <del class="diffchange diffchange-inline">Resources </del>==</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>== <ins class="diffchange diffchange-inline">See Also </ins>==</div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{USAMO box|year=1991|num-b=3|num-a=5}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{USAMO box|year=1991|num-b=3|num-a=5}}</div></td></tr>
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1=2
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Etude at 00:52, 4 July 2013
2013-07-04T00:52:23Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 00:52, 4 July 2013</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=344949#344949 Discussion on AoPS/MathLinks]</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=344949#344949 Discussion on AoPS/MathLinks]</div></td></tr>
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Etude
https://artofproblemsolving.com/wiki/index.php?title=1991_USAMO_Problems/Problem_4&diff=22401&oldid=prev
Boy Soprano II: problem and elementary solution
2008-01-13T23:04:58Z
<p>problem and elementary solution</p>
<p><b>New page</b></p><div>== Problem ==<br />
<br />
Let <math>\, a =(m^{m+1} + n^{n+1})/(m^m + n^n), \,</math> where <math>\,m\,</math> and <math>\,n\,</math> are positive integers. Prove that <math>\,a^m + a^n \geq m^m + n^n</math>.<br />
<br />
[You may wish to analyze the ratio <math>\,(a^N - N^N)/(a-N),</math> for real <math>\, a \geq 0 \,</math> and integer <math>\, N \geq 1</math>.]<br />
<br />
== Solution ==<br />
<br />
Let us assume without loss of generality that <math>m\ge n</math>. We then note that<br />
<cmath> m-a = \frac{m^{m+1} + m \cdot n^n}{m^m+n^n} - \frac{m^{m+1} - n^{n+1}}{m^m+n^n} = n^n \frac{m-n}{m^m+n^n} \qquad (*) . </cmath><br />
Similarly,<br />
<cmath> a-n = m^m \frac{m-n}{m^m+n^n} \qquad (**) . </cmath><br />
<br />
We note that equations <math>(*)</math> and <math>(**)</math> imply that <math>n \le a \le m</math>. Then <math>a/m \le 1 \le a/n</math>, so<br />
<cmath> \frac{1}{m} \sum_{i=0}^{m-1} (a/m)^i \le 1 \le \frac{1}{n} \sum_{i=0}^{n-1} (a/n)^i . </cmath><br />
Multiplying this inequality by <math>m^m n^n(m-n)/(m^m+n^n)</math>, we have<br />
<cmath> n^n \frac{(m-n)}{m^m+n^n} \sum_{i=0}^{m-1} a^i m^{m-1-i} \le m^m \frac{(m-n)}{m^m+n^n} \sum_{i=0}^{n-1} a^i n^{n-1-i} . </cmath><br />
It then follows that<br />
<cmath> \begin{align*}<br />
m^m - a^m &= (m-a) \sum_{i=0}^{m-1} a^i m^{m-1-i} = n^n \frac{m-n}{m^m+n^n} \sum_{i=0}^{m-1} a^i m^{m-1-i} \\<br />
&\le m^m \frac{m-n}{m^m+n^n} \sum_{i=0}^{n-1} a^i n^{n-1-i} = (a-n) \sum_{i=0}^{n-1} a^i n^{n-1-i} \\<br />
&= a^n - n^n .<br />
\end{align*} </cmath><br />
Rearranging this inequality, we find that <math>a^m + a^n \ge m^m + n^n</math>, as desired. <math>\blacksquare</math><br />
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{{alternate solutions}}<br />
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== Resources ==<br />
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{{USAMO box|year=1991|num-b=3|num-a=5}}<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=344949#344949 Discussion on AoPS/MathLinks]<br />
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[[Category:Olympiad Algebra Problems]]</div>
Boy Soprano II