# Question using induction to prove an inequality [duplicate]

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!

• Sorry, but I don't know how to use Latex and MathJax May 7, 2019 at 14:15

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.

• Hi, how did you get to the second line proof? May 9, 2019 at 23:01
• @Colao: please see comment. May 10, 2019 at 7:57
• yeah i've just understood now, thanks! May 11, 2019 at 2:31
• If the second line is true, then "sqrt(n+1)" will be also smaller than the LHS of first line, right? (correct me If I'm wrong, please) May 11, 2019 at 2:40
• @DanielSehnColao: right! May 11, 2019 at 12:10