How to prove this inequality using am gm? Is it true that
$$\frac{1}{3k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}>\frac{1}{2k+1}+\frac{1}{2k+2}$$
for all natural k? Is bashing going to work?
 A: You can compare the RHS and the LHS (line 3 in Jack d'Auruzio's answer) this way:
$\frac{1}{6k+2} \ $ (inside LHS),  is bigger than: $\frac{1}{6k+3}\ $  (inside RHS)
$\frac{1}{6k+4} > \frac{1}{6k+6}$ aswell.  
We will now prove that what's left still "respects" the original (weaker) inequality:
$\frac{1}{6k+2} + \frac{1}{6k+4} > \frac{1}{6k+3} + \frac{1}{6k+3} \ \ \ \  (1)$.
We could do this very easily without using AM > GM, just do the calculations:
$\frac{(6k+4)+(6k+2)}{(6k+2)(6k+4)} > \frac{(6k+3)+(6k+3)}{(6k+3)(6k+3)}$ 
$ \frac{12k+6}{36k^2+36k+8} > \frac{12k+6}{36k^2+36k+9}$, which is true for any k in N.
If you want to use AM > GM you can skip my last 2 lines and observe that:
If $AM > GM$, and since $GM > HM$, then $AM > HM$ , therefore  $\frac{2}{HM} > \frac{2}{AM}$.   Consider $6k+2$ and $6k+4$, their $\frac{2}{HM}$ is just the left side of (1), while the $\frac{2}{AM} $ is just on the right side, therefore:
$\frac{2}{HM} > \frac{2}{AM}$ leads to:
$\frac{1}{6k+2} + \frac{1}{6k+4} > \frac{1}{6k+3} + \frac{1}{6k+3}$
A: The LHS is 
$$\frac{1}{6k+2}+\frac{1}{6k+2}+\frac{1}{6k+4}+\frac{1}{6k+4}+\frac{1}{6k+6}+\frac{1}{6k+6}$$
while the RHS is
$$\frac{1}{6k+3}+\frac{1}{6k+3}+\frac{1}{6k+3}+\frac{1}{6k+6}+\frac{1}{6k+6}+\frac{1}{6k+6}$$
hence we just need to prove that:
$$\frac{1}{6k+2}+\frac{1}{6k+2}+\frac{1}{6k+4}+\frac{1}{6k+4}>\frac{1}{6k+3}+\frac{1}{6k+3}+\frac{1}{6k+3}+\frac{1}{6k+6}$$
that follows from:
$$\frac{1}{6k+2}-\frac{1}{6k+3}=\frac{1}{36k^2+30k+6},$$
$$\frac{1}{6k+4}-\frac{1}{6k+6}=\frac{2}{36k^2+60k+24},$$
$$\frac{1}{6k+3}-\frac{1}{6k+4}=\frac{1}{36k^2+42k+12},$$
$$\frac{2}{36k^2+30k+6}+\frac{2}{36k^2+60k+24}>\frac{1}{36k^2+42k+12}.\tag{3}$$
To prove $(3)$, you can use AM-GM, AM-HM or many other methods, it is not a tight inequality.

Another possibility is given by writing both terms as geometric series. Since:
$$\frac{1}{6(k+1)-\eta}=\frac{1}{6(k+1)}+\sum_{j=1}^{+\infty}\frac{\eta^j}{(6(k+1))^{j+1}}$$
and so on, we just have to prove that:
$$2\sum_{j=1}^{+\infty}\frac{1+2^{j}+4^j}{(6(k+1))^{j+1}}>3\sum_{j=1}^{+\infty}\frac{1+3^{j}}{(6(k+1))^{j+1}}$$
that is trivial since $2(1+2^j+4^j)> 3(1+3^j)$ holds by induction for any $j\geq 1$.
