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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 August 14th, 2019, 09:55 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 728 Thanks: 98 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: 6,834 Thanks: 733 $\sqrt{1}=1$, $1+\sqrt{2}\gt 2$ 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 $n\gt 2$. Thanks from idontknow Last edited by skipjack; August 14th, 2019 at 01:30 PM. August 14th, 2019, 01:46 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 The inequality is an easy consequence of √$n \geqslant 1$, where $n \geqslant 1$. Thanks from greg1313 and idontknow August 14th, 2019, 02:58 PM   #4
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Quote:
 Originally Posted by idontknow Prove that $\displaystyle \sqrt{1} +\sqrt{2} +...+\sqrt{n} \geq n$.
By induction:
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: 728 Thanks: 98 This proof also holds true . $\displaystyle s_n >\underbrace{\sqrt{1}+\sqrt{1} +... }_{n} =n\geq n$ Tags inequality, sqrt Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post DecoratorFawn82 Algebra 2 October 1st, 2017 06:03 PM StillAlive Calculus 5 September 2nd, 2016 11:45 PM nikkor180 Real Analysis 3 June 25th, 2009 03:18 AM Mathworm Math Events 11 March 3rd, 2007 04:35 PM

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