Prove the following inequality by induction for all $n\in \Bbb N$

Question

I know how to prove the base case for this. But how would I continue from there?

Answer 1

Base case is fine.

Hypothesis statement:

Suppose it is true for some $n \in \mathbb{N}$.

Then, $P(n) : \sum_{i=1}^n \frac{1}{\sqrt{i}} \geq \sqrt{n}$

Now, we have to prove this for $n+1$, given $P(n)$ is true. So, $P(n+1): \sum_{i=1}^n \frac{1}{\sqrt{i}} + \frac{1}{\sqrt{n+1}} \geq \sqrt{n} + \frac{1}{\sqrt{n+1}} = \sqrt{n+1}\left(\sqrt{\frac{n}{n+1}} + \frac{1}{\sqrt{n+1}}\right) \geq \sqrt{n+1}$

Where the last inequality holds because the term in the paranthesis is a factor greater than one. This is because $1 \geq \frac{1}{\sqrt{n+1} + \sqrt{n}}$, so

$\sqrt{n+1} + \sqrt{n} \geq 1 \Rightarrow 1 \geq \sqrt{n+1} - \sqrt{n} \Rightarrow \sqrt{n} + 1 \geq \sqrt{n+1} \Rightarrow \frac{\sqrt{n} + 1}{\sqrt{n+1}} \geq 1 $

So, true by mathematical induction.

Answer 2

Set $$S_n =\sum_{i=1}^n \frac{1}{\sqrt{i}} \geq \sqrt{n}$$ then $$S_{n+1} =\sum_{i=1}^{n+1} \frac{1}{\sqrt{i}} \geq \sqrt{n} =S_n +\frac{1}{\sqrt{n+1}}$$ Obviously $S_1\ge 1 =\sqrt1.$

If we assume that $$S_n\ge \sqrt{n}$$ then we have that $$S_{n+1} =S_n +\frac{1}{\sqrt{n+1}} \ge \sqrt{n}+\frac{1}{\sqrt{n+1}} =\frac{\sqrt{n^2+n} +1}{\sqrt{n+1}} $$

But we know that, $$n^2\le 4n^2 +4n \Longleftrightarrow n\le 2 \sqrt{n^2+n}\Longleftrightarrow n+\color{blue}{n^2+n+1} \le 2 \sqrt{n^2+n}+\color{blue}{n^2+n+1} $$

that $$(n+1)^2 = n^2 +2n +1 \le 2 \sqrt{n^2+n}+\color{blue}{n^2+n+1} = (\sqrt{n^2+n}+1)^2$$

we just prove that, $$\sqrt{n^2+n}+1\ge n+1 \Longleftrightarrow \frac{\sqrt{n^2+n} +1}{\sqrt{n+1}}\ge \sqrt{n+1}$$

Hence, $$S_{n+1}\ge \sqrt{n}+\frac{1}{\sqrt{n+1}} =\frac{\sqrt{n^2+n} +1}{\sqrt{n+1}} \ge \sqrt{n+1}$$