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August 14th, 2019, 09:55 AM   #1
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Inequality with sqrt

Prove that $\displaystyle \sqrt{1} +\sqrt{2} +...+\sqrt{n} \geq n$.
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August 14th, 2019, 01:18 PM   #2
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$\sqrt{1}=1$, $1+\sqrt{2}\gt 2$
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 $n\gt 2$.
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Last edited by skipjack; August 14th, 2019 at 01:30 PM.
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August 14th, 2019, 01:46 PM   #3
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The inequality is an easy consequence of √$n \geqslant 1$, where $n \geqslant 1$.
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August 14th, 2019, 02:58 PM   #4
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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by idontknow View Post
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
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Last edited by topsquark; August 14th, 2019 at 03:00 PM.
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August 15th, 2019, 12:31 AM   #5
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This proof also holds true .
$\displaystyle s_n >\underbrace{\sqrt{1}+\sqrt{1} +... }_{n} =n\geq n$
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