Could someone help me to prove this inequality using the induction method?
$n \in\mathbb{N}$ (Naturals)
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ ...\ +\frac{1}{\sqrt{n}}\geq\sqrt{n}$$
I did the basis step ($n=1$), but I got stuck at the second step (induction step).
All help is appreciated!
Well, the induction step is as follows for $n\geq1$:
$$\frac{1}{\sqrt 1} + \ldots + \frac{1}{\sqrt n} + \frac{1}{\sqrt {n+1}} \geq \sqrt n + \frac{1}{\sqrt {n+1}}$$ by the induction hypothesis.
Now you have to prove that for the right-hand side,
$$ \sqrt n + \frac{1}{\sqrt {n+1}}\geq \sqrt{n+1}.$$
Consider the difference of these expressions. Then $$ \sqrt n + \frac{1}{\sqrt {n+1}} - \sqrt{n+1} = \frac{\sqrt n\sqrt{n+1} + 1 - \sqrt {n+1}\sqrt{n+1}}{\sqrt{n+1}} = \frac{\sqrt n\sqrt{n+1}-n}{\sqrt{n+1}}.$$ The numerator is $\sqrt n\sqrt{n+1} - n=\sqrt{n(n+1)}-n = \sqrt{n^2+n} - n\geq 0$ and so the above expression is $\geq 0$ as required.