341For the left part the inequality for +ve reals
(a_1+a_2+...+a_n) \left(\frac{1}{a_1} + \frac{1}{a_2}+...+\frac{1}{a_n} \right) \ge n^2 will suffice
This turned up in 1 of the RMO's of WB region, I found it nice while solving, so flicked it here -
Prove the sum 11001+11002+....13001 lies between 1 and 43.
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3 Answers
62Hint: one proof is by calculus (integration)
It will lie between 1.09827 & 1.0989. (I used calculator to find the value of something though :P)
For IIT JEE this proof would suffice... but for olympiad you have to use some simple(?) algebraic logic
341341For the right part
\frac{1}{1001} + \frac{1}{1002} +...+\frac{1}{3001}
= \sum_{k=1001}^{1500} \frac{1}{k} + \sum_{k=1501}^{2000} \frac{1}{k}+\sum_{k=2001}^{2500} \frac{1}{k}+\sum_{k=2501}^{3000} \frac{1}{k} + \frac{1}{3001}
< 500 \left[\frac{1}{1000} + \frac{1}{1500} + \frac{1}{2000} + \frac{1}{2500} \right] + \frac{1}{3001}
= \frac{1}{2} + \frac{1}{3} +\frac{1}{4} + \frac{1}{5} + \frac{1}{3001} < \frac{4}{3}
