Other Idea to show an inequality $\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$

$$\begin{cases}\dfrac{1}{\sqrt 1}\geq \dfrac{1}{\sqrt n}\\+\dfrac{1}{\sqrt 2}\geq \dfrac{1}{\sqrt n}\\+\dfrac{1}{\sqrt 3}\geq \dfrac{1}{\sqrt n}\\ \vdots\\+\dfrac{1}{\sqrt n}\geq \dfrac{1}{\sqrt n}\end{cases} \\\\$$

sum of left hands is $\underbrace{\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}}$ sum of the right hands is $n\times \dfrac{1}{\sqrt n}$ so $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n} \geq n\dfrac{1}{\sqrt n}=\dfrac{\sqrt{n^2}}{\sqrt{n}}=\sqrt{n} \checkmark$$

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.

I thought it might be instructive to present a simple approach that yields a much tighter bound than requested in the OP, and that relies on nothing more than telescoping series and straightforward arithmetic. To that end, we now proceed.

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We begin with the telescoping series

$$\sum_{k=1}^{n}\left(\sqrt{k+1}-\sqrt{k}\right)=\sqrt{n+1}-1 \tag 1$$

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Inasmuch as $\sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}$, we can write $(1)$ as

$$\sum_{k=1}^{n}\left(\frac{1}{\sqrt{k+1}+\sqrt{k}}\right)=\sqrt{n+1}-1 \tag 2$$

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Then, using $\sqrt{k+1}>\sqrt k$, we have the inequality

$$\sum_{k=1}^{n}\left(\frac{1}{2\sqrt{k}}\right)>\sqrt{n+1}-1$$

from which we see that

$$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^{n}\frac{1}{\sqrt{k}}> 2(\sqrt {n+1}-1)} \tag 3$$

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Note that $(3)$ provides a much tighter bound for the summation of interest than the one requested in the OP since

$$\sum_{k=1}^n\frac{1}{\sqrt k}>2(\sqrt {n+1} -1)> \sqrt n $$

for $n\ge 2$. It's easy to see that $\sum_{k=1}^n \frac1{\sqrt k} = \sqrt n $ for $n=1$.

And we are done!

Tools Used: Telescoping Series and straightforward arithmetic