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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.

We begin with the telescoping series

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

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

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

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

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

from which we see that

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

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

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

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

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.

We begin with the telescoping series

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

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

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

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

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

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

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.

We begin with the telescoping series

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

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

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

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

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

from which we see that

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

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

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

for $n\ge 4$. It's easy to show that $\sum_{k=1}^n \frac1{\sqrt k} \ge \sqrt n $ for $n=1,2,3$ by hand.

And we are done!

Tools Used: Telescoping Series and straightforward arithmetic

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.

We begin with the telescoping series

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

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

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

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

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

from which we see that

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

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

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

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

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.

We begin with the telescoping series

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

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

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

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

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

from which we see that

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

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

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

as was to be shown!

Tools Used: Telescoping Series and straightforward arithmetic

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.

We begin with the telescoping series

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

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

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

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

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

from which we see that

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

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

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

for $n\ge 4$. It's easy to show that $\sum_{k=1}^n \frac1{\sqrt k} \ge \sqrt n $ for $n=1,2,3$ by hand.

And we are done!

Tools Used: Telescoping Series and straightforward arithmetic