Prove that $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}}\geq \sqrt{n}$ [duplicate]

Question

Anyone who can solve it or give me an idea on how to try to do it myself?

$$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}}\geq \sqrt{n}, \;\;\;n \in \mathbb{N^*}$$

Answer 1

The inequality should be the other way around. To solve it multiply both sides by $\sqrt{n}$ and note that the RHS is $n$ and on the LHS you have $n$ terms each greater than $1$.

Answer 2

If we multiply both sides by $\sqrt n$, we obtain,:

$\sqrt{n}+\sqrt{\frac{n}2}+\cdots + 1\ge n$.

Each term on the left is at least $1$, and there are $n$ of them, so the sum is greater than or equal to $n$, as desired.

Answer 3

we have $ 1<\sqrt n, \sqrt 2 < \sqrt n ... \implies \frac{1}{\sqrt 1}+\frac{1}{\sqrt 2} ...+\frac{1}{\sqrt n} \geq \frac{1}{\sqrt n}*n = \sqrt n$

Answer 4

The denominators of the terms on the left are increasing, so the smallest term on the left is $\frac1{\sqrt{n}}$. Thus

$$\frac1{1}+\frac1{\sqrt{2}}+\cdots +\frac1{\sqrt{n}}$$ $$\geq \frac1{\sqrt{n}} + \frac1{\sqrt{n}} + \cdots + \frac1{\sqrt{n}}$$ $$= n\cdot\frac1{\sqrt{n}} = \sqrt{n}$$