July 20th, 2019, 08:45 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87  Prove inequality
(1) Prove that $\displaystyle \frac{\tan(x)}{x}<2\sqrt{1x^{2}}\; $,for $\displaystyle 0<x\leq 1$. (2) Prove that $\displaystyle \frac{\sin(N)}{N} +\frac{\sin(N+1)}{N+1} +...+\frac{\sin(2N)}{2N}>\frac{1}{6} \; $ for $\displaystyle N\in \mathbb{N}$. Last edited by skipjack; July 20th, 2019 at 01:44 PM. 
July 20th, 2019, 08:59 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 
is this a brain teaser or do you need help?

July 20th, 2019, 10:13 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87  It’s not for help; I’m just working them for fun.
Last edited by skipjack; July 20th, 2019 at 01:48 PM. 
July 23rd, 2019, 03:24 PM  #4 
Member Joined: Jun 2019 From: AZ, Seattle, San Diego Posts: 30 Thanks: 21  
July 24th, 2019, 06:03 AM  #5 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87 
Yes, about inequality (2): Since $\displaystyle S_{n}=\sum \frac{\sin(n)}{n}$ is increasing, $\displaystyle N=1$ gives the minimal number of terms, which is $\displaystyle 2$. By the first two terms: $\displaystyle \frac{\sin(N)}{N}+\frac{\sin(1+N)}{1+N}>\frac{ 1}{6}$ remains to prove the whole inequality. Can we somehow continue? Last edited by skipjack; July 24th, 2019 at 10:45 PM. 

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