my question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.
Page 35. Exercise 1. Prove the following formula by induction: $$1^3+2^3+3^3+\cdots+(n-1)^3<\frac{n^4}{4}<1^3+2^3+3^3+\cdots+n^3$$
The attempt at a solution: First I do the left side of the inequality, $$1^3+2^3+3^3+\cdots+(n-1)^3<\frac{n^4}{4}$$
For $n=2$, we have $$(2-1)^3<\frac{2^4}{4},$$ $$1<4$$ which is indeed true, now we do induction step, and add $n+1$ to both sides, we get: $$1^3+2^3+3^3+\cdots+(n-1)^3+n^3<\frac{n^4}{4}+n^3,$$ now I know from earlier exercise that $$1^3+2^3+3^3+\cdots+(n-1)^3+n^3=(1+2+3+\cdots+n)^2,$$ and $$(1+2+3+\cdots+n)^2=(\frac{n(n+1)}{2})^2=(\frac{n^2+n}{2})^2=\frac{n^4+2n^3+n^2}{4}$$so$$\frac{n^4+2n^3+n^2}{4}<n^4+n^3$$this is where I am stuck, I have continued on different paths and they all seem to give nothing, please help.
Assuming $\sum_{i=1}^{n-1} i^3 < \frac{n^4}{4}$, we see that \begin{align*} \sum_{i=i}^n i^3 &= \sum_{i=1}^{n-1} i^3 + n^3 \\ &< \frac{n^4}{4} + n^3 = \frac{n^4+4n^3}{4} \\ &< \frac{n^4 + 4n^3 + 6n^2 + 4n + 1}{4} \\ &= \frac{(n+1)^4}{4}. \end{align*}