Let $A = \cfrac{1}{\cfrac{1}{2011}+\cfrac{1}{2012}+\cfrac{1}{2013}+\cfrac{1}{2014}+\cfrac{1}{2015}+\cfrac{1}{2016}}$. $B$ is the largest integer so that $B \le A$. Find $B$
I tried to use AM-GM inequality in the denominator but it is impossible
2026-03-28 15:21:25.1774711285
Let $A = \frac{1}{\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}}$.
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$$ \frac{1}{\frac{1}{2011} + \frac{1}{2011} + \frac 1{2011} + \frac{1}{2011}+\frac{1}{2011}+\frac{1}{2011}} < \frac 1{\frac{1}{2011} +\frac{1}{2012} +\frac{1}{2013} +\frac{1}{2014} +\frac{1}{2015} +\frac{1}{2016}}\\ < \frac{1}{\frac 1{2016}+\frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}} $$
LHS $= \frac {2011}{6} > 335$ , RHS $= \frac {2016} {6} = 336$