Proving $\frac{(n+1)^4}{4}+(n+1)^3\le\frac{(n+2)^4}{4}$ for all $n \ge 1$. 
$$\frac{(n+1)^4}{4}+(n+1)^3\le\frac{(n+2)^4}{4}$$

For all $n\ge 1$. I thought that I could get rid of the denominators like this:
$$(n+1)^4+4(n+1)^3\le(n+2)^4$$
Then, maybe, take $(n+1)^3$ as common factor:
$$(n+1)^3\cdot((n+1)+4)\le(n+2)^4$$
$$(n+1)^3\cdot(n+5)\le(n+2)^4$$
But I get the feeling that I'm just getting stuck. How can I further prove this?

This came up because I was doing an induction exercise. I needed to prove for all $n \ge 1$:
$$1^3+2^3+3^3+...+n^3\le \frac{(n+1)^4}{4}$$
By the hypothesis I know that
$$1^3+2^3+3^3+...+n^3+(n+1)^3\le \frac{(n+1)^4}{4}+(n+1)^3$$
So what I finally need to prove is that
$$\frac{(n+1)^4}{4}+(n+1)^3 \le \frac{(n+2)^4}{4}$$
 A: You may just expand both expressions $(n+1)^4+4(n+1)^3$ and $(n+2)^4$, the first one is $n^4 + 8n^3 + 18n^2 + 16n + 5$ and the second one is $n^4 + 8n^3 + 24n^2 + 32n + 16$. Now you subtract them. $$(n^4 + 8n^3 + 24n^2 + 32n + 16) - (n^4 + 8n^3 + 18n^2 + 16n + 5)=6n^2 + 16n + 11\geq0$$
Therefore, $$n^4 + 8n^3 + 24n^2 + 32n + 16 \geq n^4 + 8n^3 + 18n^2 + 16n + 5\implies$$
$$\implies (n+2)^4\geq (n+1)^4+4(n+1)^3$$
In fact, this is a strict inequality.
A: Hint: Put $m = n+1$, all one needs to prove is that: $ m^4 + 4 m^3 \leq (m+1)^4$. Now just expand the right hand side.
A: It is equivalent to see that $$4(n+1)^3\leq(n+2)^4-(n+1)^4=\bigg((n+1)^2+(n+2)^2\bigg)\big(2n+3 \big)\leq \bigg((n+1)^2+(n+2)^2\bigg)\big(3n+3)\leq 3(n+1)^3+3(n+1)(n+2)^2,$$
which is always true, where we have used the identity $a^4-b^4=(a^2+b^2)(a^2-b^2)$.
A: $(a+b)^4=(a^2+2ab+b^2)^2=a^2+4a^2b^2+b^4+4a^3b+4ab^3+2a^2b^2$
$$\frac{(n+2)^4}{4}=\frac{(n+1+1)^4}{4}=\frac{(n+1)^4}{4}+(n+1)^2+\frac{1}{4}+(n+1)^3+(n+1)+\frac{1}{2}(n+1)^2$$
$$\frac{(n+2)^4}{4}=\frac{(n+1)^4}{4}+(n+1)^3+\text { some thing which is positive}$$
So, $$\frac{(n+2)^4}{4}\geq\frac{(n+1)^4}{4}+(n+1)^3 \text{ for all $n\in \mathbb{N}$}$$
