Difference between revisions of "2012 Indonesia MO Problems/Problem 2"
Skill issue7 (talk | contribs) |
Skill issue7 (talk | contribs) |
||
| Line 3: | Line 3: | ||
Let <math>n\ge 3</math> be an integer, and let <math>a_2,a_3,\ldots ,a_n</math> be positive real numbers such that <math>a_{2}a_{3}\cdots a_{n}=1</math>. Prove that | Let <math>n\ge 3</math> be an integer, and let <math>a_2,a_3,\ldots ,a_n</math> be positive real numbers such that <math>a_{2}a_{3}\cdots a_{n}=1</math>. Prove that | ||
<cmath>(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.</cmath> | <cmath>(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.</cmath> | ||
| − | |||
==Solution== | ==Solution== | ||
Revision as of 07:26, 20 December 2024
Problem
Let 
be an integer, and let 
be positive real numbers such that 
. Prove that
![\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]](http://latex.artofproblemsolving.com/7/7/f/77fb719e975225b59c64f1f3aa102c98a6c25f50.png)
Solution
By AM-GM
![\[(1+a_k)^k=(\frac{1}{k-1}+\frac{1}{k-1}+\dots+\frac{1}{k-1}+n_k)^k\geq (k\sqrt[k]{\frac{a_k}{(k-1)^{k-1}}})^k=\frac{k^k\cdot a_k}{(k-1)^{k-1}}\]](http://latex.artofproblemsolving.com/f/7/2/f72d2d93438be8a03d290199ef68809c1b298f67.png)
Substituting back into our original equation we get ![\[(1+a_2)^2(1+a_3)^3\dots(1+a_n)^n\geq \frac{2^2\cdot a_2}{1^1}\frac{3^3\cdot a_3}{2^2}\dots\frac{n^n\cdot a_n}{(n-1)^{n-1}}=\frac{2^2}{1^2}\frac{3^3}{2^2}\frac{4^4}{3^3}\dots\frac{(n-1)^{n-1}}{(n-2)^{n-2}}\frac{n^n}{(n-1)^{n-1}}\cdot a_1a_2a_3\dots a_n=n^n\cdot 1=n^n\]](http://latex.artofproblemsolving.com/c/d/2/cd28781b3f890632e8c6442622fda5177b5df836.png)
however, we only proved its 
, for the equality to happen, 
for all 
, which is impossible for all of them to be so, thus the equality is impossible
See Also
| 2012 Indonesia MO (Problems) | ||
| Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Problem 2 |
| All Indonesia MO Problems and Solutions | ||
