Difference between revisions of "2025 USAJMO Problems/Problem 4"
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By Vandermonde's, | By Vandermonde's, | ||
<cmath> \frac 12\binom{\sum a_i}2=\frac 12\left(\sum\binom{a_i}2+\sum_\text{cyc}a_ia_j\right)\ge\frac 12\sum ia_i^2\ge\sum i\binom{a_i}2, </cmath>with equality at <math>a_0=0,1</math> and <math>a_i=0</math> for <math>i>0.~\square</math> | <cmath> \frac 12\binom{\sum a_i}2=\frac 12\left(\sum\binom{a_i}2+\sum_\text{cyc}a_ia_j\right)\ge\frac 12\sum ia_i^2\ge\sum i\binom{a_i}2, </cmath>with equality at <math>a_0=0,1</math> and <math>a_i=0</math> for <math>i>0.~\square</math> | ||
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~rhydon516 (sol credits to leo) | ~rhydon516 (sol credits to leo) | ||
Revision as of 17:07, 23 March 2025
Contents
Problem
Let 
be a positive integer, and let 
be nonnegative integers such that 
Prove that![\[\sum_{i=0}^n i\binom{a_i}{2}\le\frac{1}{2}\binom{a_0+a_1+\dots+a_n}{2}.\]](http://latex.artofproblemsolving.com/f/6/1/f61c0806dd4a2a4b758862a88c803ed83245d9b5.png)
Note: 
for all nonnegative integers 
.
Solution
By Vandermonde's,
![\[\frac 12\binom{\sum a_i}2=\frac 12\left(\sum\binom{a_i}2+\sum_\text{cyc}a_ia_j\right)\ge\frac 12\sum ia_i^2\ge\sum i\binom{a_i}2,\]](http://latex.artofproblemsolving.com/2/0/c/20cfb07fe15f73d145f24f5321fd262209cf5105.png)
with equality at 
and 
for 
~rhydon516 (sol credits to leo)
See Also
| 2025 USAJMO (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAJMO Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
