Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ for every natural $n$.
I've already tried to write it like this:
$$\frac{\sqrt{1}}{1} + \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} + ... + \frac{\sqrt{n}}{n} \leq 3\sqrt{n+1} - 3$$
$$\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} + ... + \frac{\sqrt{n}}{n} \leq 3\sqrt{n+1} - 4$$
but I don't know that to do next or if it's the right way prove this.

Have you tried using induction?
Let's assume the following proposition is true for $n$:
$$\sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3$$
By induction we must prove it works for the case $n=1$ and for all the cases $n+1$.
$P(1)$:
The base case is trivial, since
$$\frac{1}{\sqrt{1}} = 1 \leq 3\sqrt{2} -3 = 3(\sqrt2 - 1)$$
Since $\sqrt2 > 1$ it is easy to see why it is true.
Now we must bring our proposition $p(n) := \sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3$ to test for all $n+1$ and see if it holds.
$P(n) \Rightarrow P(n+1)$:
$$\sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3$$
$$\sum_{k=1}^{n}\frac{1}{\sqrt{k}} + (\frac{1}{\sqrt{n+1}}) \leq 3\sqrt{n+1} -3 + (\frac{1}{\sqrt{n+1}})$$
$$\sum_{k=1}^{n+1}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3 + \frac{1}{\sqrt{n+1}} \leq 3\sqrt{n+2} - 3 $$
Therefore
$$\sum_{k=1}^{n+1}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+2} - 3 $$
You can check that $3\sqrt{n+1} + \frac{1}{\sqrt{n+1}} \leq 3\sqrt{n+2}$ is true for all $n$. Hence we proved by induction that the inequality $P(n)$ is true.
