<s>[INST] 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.
 [/INST] By induction we have

*

*base case: $n=1 \implies 1 \le 3\sqrt 2-3$

*inductive step: we assume true

$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + … + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3 \tag 1$$
and we need to prove that
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + … + \frac{1}{\sqrt{n}}+ \frac{1}{\sqrt{n+1}} \stackrel{(1)}\leq 3\sqrt{n+1} - 3+\frac{1}{\sqrt{n+1}} \stackrel{?}\le 3\sqrt{n+2} - 3$$
and last inequality is true indeed
$$ 3\sqrt{n+1} +\frac{1}{\sqrt{n+1}}\le 3\sqrt{n+2} $$
$$ 3(n+1) +1\le 3\sqrt{(n+2)(n+1)} $$
$$ 3n+4\le 3\sqrt{(n+2)(n+1)} $$
$$ 9n^2+24n+16\le 9n^2+27n+18 $$
</s>