Prove that $$\sum_{k=1}^{n-1}k^{3}\leq \frac{n^{4}}{4}\leq \sum_{k=1}^{n}k^{3}$$ for all $n\geq 2$.
This is just a random exercise to improve my proof techniques. I want to show it by induction that the statement $$P(n):\sum_{k=1}^{n-1}k^{3}\leq \frac{n^{4}}{4}\leq \sum_{k=1}^{n}k^{3}$$ is true for all $n\geq 2$.
The case $n=2$ is trivial, so $P(2)$ is true. Assume that $P(m)$ is true for some $m\geq 2$. I want to show that $P(m+1)$ is true. Splitting the expression into two inequalities, we see that
$$\sum_{k=1}^{(m+1)-1}k^{3}=\sum_{k=1}^{(m-1)+1}k^{3}=\sum_{k=1}^{m-1}k^{3}+m^{3}\leq \frac{m^{4}}{4}+m^{3}<\frac{(m+1)^{4}}{4}$$ and (I am not sure about this way) $$\frac{(m+1)^{4}}{4}\leq \sum_{k=1}^{m+1}k^{3}=\sum_{k=1}^{m}k^{3}+(m+1)^{3}=\sum_{k=1}^{m-1}k^{3}+(m+1)^{3}+m^{3}$$ $$\leq \frac{m^{4}}{4}+(m+1)^{3}+m^{3}.$$ They imply $$\sum_{k=1}^{(m+1)-1}k^{3}\leq \frac{(m+1)^{4}}{4}\leq \sum_{k=1}^{m+1}k^{3},$$ which shows that $P(m+1)$ is true. Hence $P(n)$ is true for all $n\geq 2$.
In the first direction, one should prove that $\frac{m^4}{4}+m^3<\frac{(m+1)^4}{4}$ explicitly. This, however is easy because $$ \frac{m^4}{4}+m^3=\frac{m^4+4m^3}{4} $$ and $$ \frac{(m+1)^4}{4}=\frac{m^4+4m^3+6m^2+4m+1}{4}. $$ So, as long as $m>0$, $\frac{6m^2+4m+1}{4}>0$ and so adding $\frac{m^4}{4}+m^3$ to both sides gives the desired inequality.
In the second direction, you start with the conclusion, which makes the proof circular. Instead, start with $$ \sum_{k=1}^{m+1}k^3=\sum_{k=1}^mk^3+(m+1)^3. $$ From the inductive hypothesis, you know a lower bound on the first sum. Therefore, $$ \sum_{k=1}^{m+1}k^3=\sum_{k=1}^mk^3+(m+1)^3\geq\frac{m^4}{4}+(m+1)^3\stackrel{want}{\geq}\frac{(m+1)^4}{4}. $$ All the inequalities have been proven, except for the one labeled want (and if the want were true, then the desired inequality would follow).
Now, you just have to use the standard inequality template to prove the wanted inequality (start with something you know and manipulate it until you reach the desired inequality).