Other Idea to show an inequality $\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$ $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$$
I want to prove this by Induction 
$$n=1 \checkmark\\
n=k \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}\geq \sqrt k\\
n=k+1 \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt {k+1}}\geq \sqrt {k+1}$$ so $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}+\dfrac{1}{\sqrt {k+1}}\geq \sqrt k+\dfrac{1}{\sqrt {k+1}}$$now we prove that $$\sqrt k+\dfrac{1}{\sqrt {k+1}} >\sqrt{k+1} \\\sqrt{k(k+1)}+1 \geq k+1 \\ k(k+1) \geq k^2 \\k+1 \geq k \checkmark$$ and the second method like below , 
and I want to know is there more Idia to show this proof  ? forexample  combinatorics proofs  , or using integrals ,or fourier series ,....  
Is there a close form for this summation ?
any help will be appreciated .

Integrals:
$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^{n+1}\frac1{\sqrt x}\ dx=2\sqrt{n+1}-2$$
And it's very easy to check that
$$2\sqrt{n+1}-2\ge\sqrt n$$
for $n\ge2$.
A visuallization of this argument:

From the red lines down, that area represents a sum.  From the blue line down, that represents an integral.  Clearly, the integral is smaller than the sum.
