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

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\begin{align*}
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\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{align*}
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\end{math}

Revision as of 08:45, 26 March 2019

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