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...")
 
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\begin{align*}
 
\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)})+\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*}
 
\end{align*}

Revision as of 08:19, 26 March 2019

[code] a11+a1+a2(1+a1)(1+a2)++an(1+a1)(1+a2)(1+an)=(111+a1)+(11+a11(1+a1)(1+a2))++(1(1+a1)(1+a2)(1+an1)1(1+a1)(1+a2)(1+an))=11(1+a1)(1+a2)(1+an)11(21a1)(21a2)(21an)=112n=2n12n