Difference between revisions of "2014 Canadian MO Problems/Problem 1"
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| − | \begin{ | + | \begin{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} | \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} | ||
| − | \end{ | + | \end{math} |
Revision as of 08:45, 26 March 2019
[code] \begin{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} \end{math}
