Inequality proof using real numbers with describe How to prove this inequality?
$$ 1+\left(\frac {1}{\sqrt 2}\right)+\left(\frac {1}{\sqrt 3}\right)+ \cdots +\left(\frac {1}{\sqrt n}\right)\gt 2(\sqrt{n+1}-1)$$
 A: $$ \sqrt{k+1}-\sqrt{k} = \frac{1}{\sqrt{k+1}+\sqrt{k}} < \frac{1}{2\sqrt{k}} $$
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.
A: 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.
A: 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 $n\in \Bbb{N}$.
