Difference between revisions of "2012 Indonesia MO Problems/Problem 2"

Latest revision as of 07:35, 20 December 2024

Problem

Let $n\ge 3$ be an integer, and let $a_2,a_3,\ldots ,a_n$ be positive real numbers such that $a_{2}a_{3}\cdots a_{n}=1$ . Prove that $(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.$

Solution

By AM-GM $(1+a_k)^k=(\frac{1}{k-1}+\frac{1}{k-1}+\dots+\frac{1}{k-1}+n_k)^k\geq (k\sqrt[k]{\frac{a_k}{(k-1)^{k-1}}})^k=\frac{k^k\cdot a_k}{(k-1)^{k-1}}$ Substituting back into our original equation we get $(1+a_2)^2(1+a_3)^3\dots(1+a_n)^n\geq \frac{2^2\cdot a_2}{1^1}\frac{3^3\cdot a_3}{2^2}\dots\frac{n^n\cdot a_n}{(n-1)^{n-1}}=\frac{2^2}{1^2}\frac{3^3}{2^2}\frac{4^4}{3^3}\dots\frac{(n-1)^{n-1}}{(n-2)^{n-2}}\frac{n^n}{(n-1)^{n-1}}\cdot a_1a_2a_3\dots a_n=n^n\cdot 1=n^n$ however, we only proved its $\geq n^n$ , for the equality to happen, $a_k=\frac{1}{k-1}$ for all $k$ , which is impossible for all of them to be so, thus the equality is impossible

@@ Line 1: / Line 1: @@
 ==Problem==
-Show that for any positive integers <math>a</math> and <math>b</math>, the number <cmath>n=\mathrm{LCM}(a,b)+\mathrm{GCD}(a,b)-a-b</cmath> is an even non-negative integer.
+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>
 ==Solution==
@@ Line 11: / Line 12: @@
 ==See Also==
-{{Indonesia MO box|year=2012|before=First Problem|num-a=2}}
+{{Indonesia MO box|year=2012|num-b=1|num-a=3}}
 [[Category:Intermediate Number Theory Problems]]

2012 Indonesia MO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 3
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Difference between revisions of "2012 Indonesia MO Problems/Problem 2"

Latest revision as of 07:35, 20 December 2024

Problem

Solution

See Also