$\sum _{j=1}^n\frac{1}{\sqrt{j}}\ge 2\cdot \sqrt{n+\ 1}-2$

Question

How to prove that

$\sum _{j=1}^n\frac{1}{\sqrt{j}}\ge 2\cdot \sqrt{n+\ 1}-2$

$n\ \varepsilon \ \mathbb{N}$

is valid? Using induction.

n=1

$\sum _{j=1}^1\frac{1}{\sqrt{j}}\ge 2\cdot \sqrt{1+\ 1}-2$

$\frac{1}{\sqrt{1}}\ge 2\cdot \sqrt{1+\ 1}-2$

$1\ge 2\cdot \left(\sqrt{2}-1\right)$

which is true

now n+1

$\sum _{j=1}^{n+1}\frac{1}{\sqrt{j}}\ge 2\cdot \sqrt{n+1+\ 1}-2$

$\sum _{j=1}^n\frac{1}{\sqrt{j}}+\ \frac{1}{\sqrt{n+1}}\ \ge 2\cdot \sqrt{n+2}-2$

now edit left side

$\sum _{j=1}^n\frac{1}{\sqrt{j}}+\ \frac{1}{\sqrt{n+1}}$ $\ge 2\cdot \sqrt{n+\ 1}-2\ +\ \frac{1}{\sqrt{n+1}}$

how get this to $2\cdot \sqrt{n+2}-2$ ?