Shortcut to proving inequality with sum of cubes

[Benny]

Hello everyone, could someone please help me out with the following question?

Q. Prove that the following inequality holds for all natural numbers n and rewrite the inequality using summation notation.

$$1^3 + 2^3 + ... + \left( {n - 1} \right)^3 < \frac{1}{4}n^4 < 1^3 + 2^3 + ... + n^3$$

For the summation notation part I got: $$\sum\limits_{k = 1}^n {\left( {k - 1} \right)^3 } < \frac{1}{4}n^4 < \sum\limits_{k = 1}^n {k^3 }$$

I managed to prove that the inequality is true. However, there seems to be some sort of symmetry in the inequality. Is there a 'shortcut' to this question or does it need to be done the way that I did? That is, showing separately that the left and right parts of the inequality are true and then combining the results. I'd like to know partly because I want to enhance my understanding of questions of this type. Any ehlp appreciated.

 

[Timbuqtu]

If you want to approach it analytically:
Notice that the left-hand side is a lower sum and the right-hand side is an upper sum of the integral
$$\int_0^{n}x^3 dx$$

This directly yields:

$$1^3 + 2^3 + ... + \left( {n - 1} \right)^3 \leq \frac{1}{4}n^4 \leq 1^3 + 2^3 + ... + n^3$$

It is not difficult to show that equality is not possible.

 

[Benny]

Thank you for your help Timbuqtu.