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August 14th, 2019, 09:55 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87  Inequality with sqrt
Prove that $\displaystyle \sqrt{1} +\sqrt{2} +...+\sqrt{n} \geq n$.

August 14th, 2019, 01:18 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716 
$\sqrt{1}=1$, $1+\sqrt{2}\gt 2$ For $n\gt 2$, $\displaystyle\sum_{k=1}^n\sqrt{k}\gt \int_0^n \sqrt{x}dx=\frac{2}{3}n^{\frac{3}{2}}\gt n$ The last step results from $\frac{2\sqrt{n}}{3}\gt 1$ for $n\gt 2$. Last edited by skipjack; August 14th, 2019 at 01:30 PM. 
August 14th, 2019, 01:46 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
The inequality is an easy consequence of √$n \geqslant 1$, where $n \geqslant 1$.

August 14th, 2019, 02:58 PM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 927 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
n = 1: $\displaystyle \sqrt{1} \geq1$? Check. Assume that this is true for some value of k = n. Then we can say $\displaystyle \sum_{i = 1}^k \sqrt{i} \geq k$ Then we wish to show that the inequality is true for n = k + 1. $\displaystyle \sum_{i = 1}^{k + 1} \sqrt{i} = \sum_{i = 1}^k \sqrt{i} + \sqrt{k + 1} \geq k + \sqrt{k + 1}$ by hypothesis. So if we can show that $\displaystyle \sum_{i = 1}^{k + 1} \sqrt{i} \geq k + \sqrt{k + 1} \geq k + 1$ we are done. Is $\displaystyle k + \sqrt{k + 1} ~ \text{?} ~ k + 1$ (The "?" stands for "not less than." I don't know how to do that one in LaTeX.) $\displaystyle \sqrt{k + 1} ~ \text{?} ~ 1$. Now, we know that $\displaystyle \sqrt{k + 1} \geq 1$ for all positive integer k. Therefore we are done. Dan Last edited by topsquark; August 14th, 2019 at 03:00 PM.  
August 15th, 2019, 12:31 AM  #5 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87 
This proof also holds true . $\displaystyle s_n >\underbrace{\sqrt{1}+\sqrt{1} +... }_{n} =n\geq n$ 

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