Difference between revisions of "1991 AIME Problems/Problem 15"
Gabiloncho (talk | contribs) (→Solution) |
Gabiloncho (talk | contribs) (→Solution) |
||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
− | We start by recalling the following simple inequality: Let <math>a_{}^{}</math> and <math>b_{}^{}</math> denote two | + | We start by recalling the following simple inequality: Let <math>a_{}^{}</math> and <math>b_{}^{}</math> denote two real numbers, then <math>\sqrt{a_{}^{2}+b_{}^{2}}\geq (a+b)/\sqrt{2}</math>, with equality if and only if <math>a_{}^{}=b_{}^{}</math>. Applying this inequality to the given sum, one has |
<math> | <math> | ||
\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}}\, , | \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>. | + | |
+ | 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>. Therefore, <math>S_{n}\geq(n^{2}+t)/\sqrt{2}</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 17:49, 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 real numbers, then , with equality if and only if . Applying this inequality to the given sum, one has
where we have used the well-known fact that , and we have defined . Therefore, .
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 |