2012 Indonesia MO Problems/Problem 2
Revision as of 02:46, 20 December 2024 by 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...")
By AM-GM,
\[(1+a_k)^k=(\frac{1}{k-1}+\frac{1}{k-1}+\dots+\frac{1}{k-1}+n_k)^k\geq (k\sqrt[k]{\frac{a_k}{(k-1)^{k-1}})^k=\frac{k^k\cdot a_k}{(k-1)^{k-1}}\] (Error compiling LaTeX. Unknown error_msg)
Substituting back into our original equation we get ![\[(1+a_2)^2(1+a_3)^3\dots(1+a_n)^n\geq \frac{2^2cdot a_2}{1^1}\frac{3^3\cdot a_3}{2^2}\dots\frac{n^n\cdot a_n}{(n-1)^{n-1}}=\frac{2^2}{1^2}\frac{3^3}{2^2}\frac{4^4}{3^3}\dots\frac{(n-1)^{n-1}}{(n-2)^{n-2}}\frac{n^n}{(n-1)^{n-1}}\cdot a_1a_2a_3\dots a_n=n^n\cdot 1=n^n\]](http://latex.artofproblemsolving.com/a/6/8/a682bb6361e0033fc47e907e5ba33257ed612f44.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
