Difference between revisions of "2012 Indonesia MO Problems/Problem 2"
Skill issue7 (talk | contribs) (Created page with "By AM-GM, <cmath>(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}}</cmath> Substitu...") |
Skill issue7 (talk | contribs) |
||
| Line 1: | Line 1: | ||
By AM-GM, | By AM-GM, | ||
| − | <cmath>(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}}</cmath> | + | <cmath>(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}}</cmath> |
| − | Substituting back into our original equation we get <cmath>(1+a_2)^2(1+a_3)^3\dots(1+a_n)^n\geq \frac{2^ | + | Substituting back into our original equation we get <cmath>(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</cmath> |
however, we only proved its <math>\geq n^n</math>, for the equality to happen, <math>a_k=\frac{1}{k-1}</math> for all <math>k</math>, which is impossible for all of them to be so, thus the equality is impossible | however, we only proved its <math>\geq n^n</math>, for the equality to happen, <math>a_k=\frac{1}{k-1}</math> for all <math>k</math>, which is impossible for all of them to be so, thus the equality is impossible | ||
Revision as of 02:50, 20 December 2024
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
