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How to prove this inequality?

$$ 1+\left(\frac {1}{\sqrt 2}\right)+\left(\frac {1}{\sqrt 3}\right)+ \cdots +\left(\frac {1}{\sqrt n}\right)\gt 2(\sqrt{n+1}-1)$$

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hint - think riemann sum... – peter a g Mar 2 at 18:03
    
Are you allowed to use integrals ? – Yves Daoust Mar 2 at 18:03
    
No , not allowed – matinhabi Mar 2 at 18:05
up vote 9 down vote accepted

$$ \sqrt{k+1}-\sqrt{k} = \frac{1}{\sqrt{k+1}+\sqrt{k}} < \frac{1}{2\sqrt{k}} $$ hence by summing these inequalities for $k=1,2,\ldots,n$ we get: $$ \sqrt{n+1}-\sqrt{1} < \frac{1}{2}\left(\frac{1}{\sqrt{1}}+\ldots+\frac{1}{\sqrt{n}}\right)$$ as wanted.

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Wow!!! It's very great. Thanks very much. – matinhabi Mar 2 at 18:15

You can do this by induction.

The base case is easy.

By induction it suffices to show that $2(\sqrt{n+1}-1)+\frac{1}{\sqrt{n+1}} \ge 2(\sqrt{n+2}-1)$. Rearranging this, we see that this is equivalent to $\frac{1}{\sqrt{n+1}} \ge 2(\sqrt{n+2}-\sqrt{n+1})$. We can rewrite the right-hand side of this using the formula $a^2-b^2=(a+b)(a-b)$: so the above is equivalent to $\frac{1}{\sqrt{n+1}} \ge \frac{2[(n+2)-(n+1)]}{\sqrt{n+2}+\sqrt{n+1}}$, i.e. $\frac{1}{\sqrt{n+1}} \ge \frac{2}{\sqrt{n+2}+\sqrt{n+1}}$. It should be easy now to see why this inequality holds, and so we're done.

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In the proof below, $n \in \Bbb{N}$ is an arbitrary integer.

Lemma 1:
$$ \frac{1}{\sqrt{n+1}} > 2\sqrt{n+2} -2\sqrt{n+1} $$ Proof of Lemma 1: $$ 9>8\\ 9+12n+4n^2 > 8+12n+4n^2\\ (3+2n)^2 > 4(n+1)(n+2)\\ 3+2n > 2\sqrt{(n+1)(n+2)}\\ 1 > 2\sqrt{(n+1)(n+2)}-2(n+1)\\ \frac{1}{\sqrt{n+1}} > 2\sqrt{n+2} -2\sqrt{n+1} $$ Corollary 2: $$ \frac{1}{\sqrt{n+1}} > 2(\sqrt{n+2}-1) -2(\sqrt{n+1}-1) $$ We will use this corollary to help establish the relationship $$ 1+\left(\frac {1}{\sqrt 2}\right)+\left(\frac {1}{\sqrt 3}\right)+ \cdots +\left(\frac {1}{\sqrt n}\right)\gt 2(\sqrt{n+1}-1)$$ by induction. Basis ($n=1$: $$ 2(\sqrt{1+1} - 1) = 2\sqrt{2} -2 ~0.83 $$ Now assume the relationship holds for some value $n$ -- we show that it also holds for $n+1$. Proof: $$ \left[ 1+\left(\frac {1}{\sqrt 2}\right)+\left(\frac {1}{\sqrt 3}\right)+ \cdots +\left(\frac {1}{\sqrt n}\right)\right] + \left(\frac{1}{\sqrt{n+1}}\right) \\ > 2(\sqrt{n+1}-1) + \frac{1}{\sqrt{n+1}} > 2(\sqrt{n+1}-1) + 2(\sqrt{n+2}-1) -2(\sqrt{n+1}-1) = 2(\sqrt{(n+1)+1}-1) $$ so the relationship holds for $n+1$. This establishes induction, so the relationship hoolds for all $n\in \Bbb{N}$.

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