You know that ${\displaystyle \sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}}$, and your goal is to show that ${\displaystyle \sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}\ge\sqrt{n+1}}}$. Observe that
$$ \sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}} = \sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}}} + {1 \over \sqrt{n+1}}$$
$$\geq \sqrt{n} + {1 \over \sqrt{n+1}}$$
You use the induction hypothesis in the above line. So what you need to show is
$$\sqrt{n} + {1 \over \sqrt{n+1}} \geq \sqrt{n+1}$$
At this point you can basically try to fool around with the algebra to get it to work out. One example of this would be to multiply both sides by $\sqrt{n+1}$, getting
$$\sqrt{n(n+1)} + 1 \geq n + 1$$
Or equivalently,
$$\sqrt{n(n+1)} \geq n$$
Squaring both sides gives
$$n^2 + n \geq n^2$$
This last equation is obviously true. To make the argument rigorous, you just observe that these steps are reversible; going in the opposite direction from above takes you from $n^2 + n \geq n^2$ to $\sqrt{n} + {1 \over \sqrt{n+1}} \geq \sqrt{n+1}$.
Some people might not like doing this sort of reversal-of-steps argument, but it does have an advantage that you don't really have to see anything clever to do it; ususally playing around with the algebra enough will eventually lead to something obvious.