Prove by induction that $ 1^2+2^2+...+(n-1)^2In Apostol's «Calculus I» on page 33 there is the following proof by induction:

To prove:
  $$
1^2+2^2+...+(n-1)^2<n^3/3<1^2+2^2+...+n^2
$$
  Solution:
  Consider the leftmost iequality first, and left us refer to this formula as $A(n)$. It is easy to verify this assertion directly for the first few values of n. Thus, for example, when n takes the values 1, 2, 3, the assertion is true.
  Our object is to prove that $A(n)$ is true for every positive integer n. The procedure is as follows: Assume the assertion has been proved for a particular value of n, say for $n = k$. That is, assume we have proved
  $$
A(k): 1^2+2^2+…+(k-1)^2<\frac{k^3}{3}
$$
  for a fixed $k\geqslant 1$. Now using this, we shall deduce the corresponding result for $k+1$:
  $$
A(k+1): 1^2+2^2+…+k^2<\frac{(k+1)^3}{3}
$$
  Start with $A(k)$ and add $k^2$ to both sides. This gives the inequality
  $$
1^2+2^2+…+k^2<\frac{k^3}{3}+k^2
$$
  To obtain $A(k+1)$ as a consequence of this, it suffices to show that
  $$
\frac{k^3}{3}+k^2<\frac{(k+1)^3}{3}
$$

So my question is why does it suffice to to show the last inequality is true to finish the proof by induction? I don't see how is $A(k+1)$ follows from $A(k)$. It is clear to me that
$$
A(k) + k^2 < A(k+1)
$$
But it is unclear why does it make sense in this case.

If $a<b$ and $b\le c$, then $a<c$.
