In Apostol's One-Variable Calculus, with an Introduction to Linear Algebra, when discussing the method of exhaustion for solving for the area under a curve (specifically $x^2$), Apostol sets up the following inequality:
$$1^2 + 2^2 + ... + (n-1)^2 < \frac{n^3}{3} < 1^2 + 2^2 + ... + n^2$$
He goes on to state that it is valid for every integer $n\geq1$ and that they can be deduced easily as consequences of the following formulas:
1.3) $$1^2 + 2^2 + ... + n^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$$
1.4) $$1^2 + 2^2 + ... + (n-1)^2 = \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}$$
I'm a bit confused at how he is able to deduce this from just 1.3 and 1.4 alone. Is it because since $n\geq1$, $\frac{n^2}{2}$ is certainly greater than $\frac{n}{6}$ and so we know 1.3 is greater than $\frac{n^3}{3}$ and vice versa for 1.4? Is there a more concrete explanation for how 1.3 and 1.4 setup the inequality? An induction proof is shared later on but I'm specifically interested in the intuition of the above!
RHS of 1.3) is greater than $n^3/3$, this is really obvious because the RHS is just $n^3/3$ plus something that is positive. That something is $(n^2/2 + n/6)$.
RHS of 1.4) is smaller than $n^3/3$. OK, this is not so obvious but still quite obvious. That's true because $$(-n^2/2 + n/6) \lt 0$$ for every natural number $n$. You can prove this easily yourself e.g. by induction. Or, if you don't want to use induction (to prove this inequality), you can just notice that it is equivalent to $$n(3n-1)\gt0$$ which of course is true for every natural number $n$.
Note: RHS means "right hand side"
These two observations prove that the sum in 1.3) is greater than $n^3/3$ while the sum in 1.4) is smaller than $n^3/3$, and that is exactly what you want to prove here.