How can I prove this inequality for every natural number $\geq 1$?

Question

I've to resolve a inequality using the induction: for every natural number $\geq 1$

$$ \sum_{k=1}^n \frac{1}{\sqrt{k}} \geq \sqrt{n} $$

At the end, I arrive at this result, but I don't know how I can continue: $\sqrt{n}+1/\sqrt{n+1} \geq \sqrt{n+1}$

Also: https://math.stackexchange.com/q/2149448/42969, https://math.stackexchange.com/q/1371267/42969. — Martin R, Sep 11 '17 at 13:07
I have removed ([tag:proof-verification]) tag, since the post actually contains no proof/solution which is supposed to be checked. — Martin Sleziak, Sep 22 '17 at 16:38

score 5 · Answer 1 · answered Sep 11 '17 at 12:51

5

it is $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}>n\frac{1}{\sqrt{n}}=\sqrt{n}$$

answered Sep 11 '17 at 12:51

sorry, i don't have understand... – fenigo69 Sep 11 '17 at 12:54
since $$\frac{1}{\sqrt{n}}$$ is the smallest summand – Dr. Sonnhard Graubner Sep 11 '17 at 12:56
ok, but sorry, i don't understand how can i proove the inequation using that – fenigo69 Sep 11 '17 at 13:01
we get $n$ summands and $$\frac{1}{\sqrt{n}}$$ is the smallest so your sum is graeter than $$n\cdot \frac{1}{\sqrt{n}}$$ – Dr. Sonnhard Graubner Sep 11 '17 at 13:05
ahhh ok,i got it, thanks a lot – fenigo69 Sep 11 '17 at 13:08
you are welcome! – Dr. Sonnhard Graubner Sep 11 '17 at 13:09
there is a discussion about this answer at math.meta – miracle173 Sep 22 '17 at 16:38

1 Answers1