Difference between revisions of "2014 Canadian MO Problems/Problem 1"

(Created page with "\begin{align*} \frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots (1+a_n)}\\=(1-\frac{1}{1+a_1})+(\frac{1}{1+a_1}-\frac{1}{(1+a_1)(1+a_2)})+\c...")
 
(4 intermediate revisions by the same user not shown)
Line 1: Line 1:
\begin{align*}
+
== Problem 1 ==
\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots (1+a_n)}\\=(1-\frac{1}{1+a_1})+(\frac{1}{1+a_1}-\frac{1}{(1+a_1)(1+a_2)})+\cdots+(\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_{n-1})}-\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_n)})\\=1-\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_n)}\\\geq 1-\frac{1}{(2\sqrt{1\cdot a_1)}(2\sqrt{1\cdot a_2)}\cdots (2\sqrt{1\cdot a_n)}}\\=1-\frac{1}{2^n}\\=\frac{2^n-1}{2^n}
+
Let <math>a_1,a_2,\dots,a_n</math> be positive real numbers whose product is <math>1</math>. Show that the sum <math>\textstyle\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\frac{a_3}{(1+a_1)(1+a_2)(1+a_3)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}</math> is greater than or equal to <math>\frac{2^n-1}{2^n}</math>.
\end{align*}
+
== Solution ==
 +
<math>\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots (1+a_n)}\\=(1-\frac{1}{1+a_1})+(\frac{1}{1+a_1}-\frac{1}{(1+a_1)(1+a_2)})+\cdots+(\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_{n-1})}-\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_n)})\\=1-\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_n)}\\\geq 1-\frac{1}{(2\sqrt{1\cdot a_1)}(2\sqrt{1\cdot a_2)}\cdots (2\sqrt{1\cdot a_n)}}\\=1-\frac{1}{2^n}\\=\frac{2^n-1}{2^n}</math>

Revision as of 11:54, 28 March 2019

Problem 1

Let $a_1,a_2,\dots,a_n$ be positive real numbers whose product is $1$. Show that the sum $\textstyle\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\frac{a_3}{(1+a_1)(1+a_2)(1+a_3)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}$ is greater than or equal to $\frac{2^n-1}{2^n}$.

Solution

$\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots (1+a_n)}\\=(1-\frac{1}{1+a_1})+(\frac{1}{1+a_1}-\frac{1}{(1+a_1)(1+a_2)})+\cdots+(\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_{n-1})}-\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_n)})\\=1-\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_n)}\\\geq 1-\frac{1}{(2\sqrt{1\cdot a_1)}(2\sqrt{1\cdot a_2)}\cdots (2\sqrt{1\cdot a_n)}}\\=1-\frac{1}{2^n}\\=\frac{2^n-1}{2^n}$