1991 AIME Problems/Problem 15: Difference between revisions
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<math> | <math> | ||
\sum_{k=1}^{n}\sqrt{(2k-1)^{2}+a_{k}^{2}}\geq \frac{1}{\sqrt{2}}\sum_{k=1}^{n}[(2k-1)^{2}+ | \sum_{k=1}^{n}\sqrt{(2k-1)^{2}+a_{k}^{2}}\geq \frac{1}{\sqrt{2}}\sum_{k=1}^{n}[(2k-1)+a_{k}]=\frac{n^{2}+t}{\sqrt{2}}\, , | ||
</math> | </math> | ||
where we have used the well-known fact that <math>\sum_{k=1}^{n}(2k-1)=n^{2}</math>, and we have defined <math>t=\sum_{k=1}^{n}a_{k}</math>. | |||
== See also == | == See also == | ||
{{AIME box|year=1991|num-b=14|after=Last question}} | {{AIME box|year=1991|num-b=14|after=Last question}} | ||
Revision as of 17:46, 19 April 2007
Problem
For positive integer 
, define 
to be the minimum value of the sum
where 
are positive real numbers whose sum is 17. There is a unique positive integer 
for which is also an integer. Find this .
Solution
We start by recalling the following simple inequality: Let 
and 
denote two positive real numbers, then 
, with equality if and only if 
. Applying this inequality to the given sum, one has
![$\sum_{k=1}^{n}\sqrt{(2k-1)^{2}+a_{k}^{2}}\geq \frac{1}{\sqrt{2}}\sum_{k=1}^{n}[(2k-1)+a_{k}]=\frac{n^{2}+t}{\sqrt{2}}\, ,$](http://latex.artofproblemsolving.com/8/8/6/886a6dbd7990337fafe4b9aa869d88d614858f5f.png)
where we have used the well-known fact that 
, and we have defined 
.
See also
| 1991 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
