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 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 . $\displaystyle s_n >\underbrace{\sqrt{1}+\sqrt{1} +... }_{n} =n\geq n$

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