Cubic Sums Bound a Quartic Term

This problem explores a classical sandwich inequality involving sums of cubes and a quartic term. By splitting the double inequality into two separate statements, we can apply both direct algebraic manipulation and mathematical induction—two fundamental proof techniques in the HSC Extension 1 and 2 syllabuses. Working through this problem will strengthen your ability to handle inequalities, manipulate sum formulas, and construct clean inductive arguments.

Problem Statement

Prove by induction that for all integers $n \geq 2$:

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

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Hints

This is a "sandwich" inequality with two parts:

• LHS inequality: $\sum_{k=1}^{n-1} k^3 < \frac{n^4}{4}$ (needs induction)
• RHS inequality: $\frac{n^4}{4} < \sum_{k=1}^{n} k^3$ (can be proven directly)

Recall: $\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}$

For the LHS induction step, you'll need to show $\frac{k^4}{4} + k^3 < \frac{(k+1)^4}{4}$, which simplifies to $0 < 6k^2 + 4k + 1$.

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Solutions

Part 1: RHS Inequality (Direct Proof)

Show $\frac{n^4}{4} < \sum_{k=1}^{n} k^3$ for $n \geq 2$.

Because the sum of the first $n$ cubes has a simple closed form, we can rewrite the RHS and compare it to the quartic term directly. Using the sum formula:

$$\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4} = \frac{n^4 + 2n^3 + n^2}{4}$$

We need: $\frac{n^4}{4} < \frac{n^4 + 2n^3 + n^2}{4}$

This simplifies to: $0 < 2n^3 + n^2 = n^2(2n + 1)$

Since $n \geq 2 > 0$, this is clearly true. RHS proven. \checkmark

Part 2: LHS Inequality (Induction)

Prove $\sum_{k=1}^{n-1} k^3 < \frac{n^4}{4}$ for $n \geq 2$.

Base case ($n=2$):

$$\sum_{k=1}^{1} k^3 = 1^3 = 1 < \frac{2^4}{4} = 4 \quad \checkmark$$

Inductive hypothesis: Assume $\sum_{k=1}^{m-1} k^3 < \frac{m^4}{4}$ for some $m \geq 2$.

Inductive step: Prove $\sum_{k=1}^{m} k^3 < \frac{(m+1)^4}{4}$.

$$\begin{aligned} \sum_{k=1}^{m} k^3 &= \sum_{k=1}^{m-1} k^3 + m^3 \\ &< \frac{m^4}{4} + m^3 \quad \text{(by IH)} \\ &= \frac{m^4 + 4m^3}{4} \end{aligned}$$

To complete the induction, we must verify that this bound is still less than the target expression for $n = m+1$, i.e. $\frac{(m+1)^4}{4}$. Need to show: $\frac{m^4 + 4m^3}{4} < \frac{(m+1)^4}{4}$

Equivalently: $m^4 + 4m^3 < (m+1)^4$

Expand RHS:

$$(m+1)^4 = m^4 + 4m^3 + 6m^2 + 4m + 1$$

So we need: $m^4 + 4m^3 < m^4 + 4m^3 + 6m^2 + 4m + 1$

Simplifies to: $0 < 6m^2 + 4m + 1$

This is true for all $m \geq 2$. \checkmark

By induction, LHS proven. Combining both parts, the full sandwich inequality holds. $\blacksquare$

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Takeaways

• The sandwich inequality can be split into two separate inequalities; the lower bound is proved by induction while the upper bound can be proved directly using the sum of cubes formula.
• The sum of cubes formula, $\sum k^3 = [n(n+1)/2]^2$, is a powerful tool that often simplifies inequality proofs.
• In the induction step, it is crucial to express $\sum_{k=1}^{m} k^3$ as $\sum_{k=1}^{m-1} k^3 + m^3$ and then use the inductive hypothesis to bound the sum, reducing the inequality to a polynomial comparison.
• Always verify the base case and check that every algebraic manipulation is valid for the given range of $n$.

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Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Distributions, HSC Complex Numbers