Power Inequality via AM-GM and Pairing

This problem is a classic Extension 2 inequality that rewards careful algebraic preparation. Rather than attacking the original inequality directly, the three-part structure guides you to factorise both sides, divide through by a positive quantity to reduce the problem, and then apply the AM-GM inequality to a geometric series. An elegant alternative—pairing symmetric terms—is also developed.

Problem Statement

Let $a$ be a real number such that $a > 1$, and let $n$ be an integer where $n \ge 2$.

i. Factorise $a^n - 1$ and show that $n\!\left(a^{\frac{n+1}{2}} - a^{\frac{n-1}{2}}\right) = n a^{\frac{n-1}{2}}(a - 1)$.

ii. Hence, show that the inequality $a^n - 1 > n\!\left(a^{\frac{n+1}{2}} - a^{\frac{n-1}{2}}\right)$ is equivalent to proving:

$$1 + a + a^2 + \dots + a^{n-1} > n\, a^{\frac{n-1}{2}}$$

iii. By applying the Arithmetic Mean–Geometric Mean (AM-GM) inequality, or otherwise, prove that $a^n - 1 > n\!\left(a^{\frac{n+1}{2}} - a^{\frac{n-1}{2}}\right)$.

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Hints

• Part (i): Recall the standard factorisation for the difference of two $n$-th powers: $x^n - y^n$. Apply this with $y = 1$. For the right-hand side, look for the highest common factor you can extract from the indices.
• Part (ii): Equate your expanded LHS to the factorised RHS from Part (i). Since you are given $a > 1$, what do you know about the sign of $a - 1$? How does this affect the inequality sign if you divide both sides by it?
• Part (iii): How many terms are in the series $1 + a + a^2 + \dots + a^{n-1}$? Set up the AM-GM inequality for these specific terms. To find the geometric mean, use the formula for the sum of an arithmetic progression to calculate the exponent of the product: $0 + 1 + 2 + \dots + (n-1)$.

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Solutions

Part (i)

Using the standard polynomial factorisation:

$$a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + \dots + a + 1)$$

For the right-hand side, factor out $a^{\frac{n-1}{2}}$:

$$n\!\left(a^{\frac{n+1}{2}} - a^{\frac{n-1}{2}}\right) = n\, a^{\frac{n-1}{2}}\!\left(a^{\frac{n+1}{2} - \frac{n-1}{2}} - 1\right) = n\, a^{\frac{n-1}{2}}(a - 1)$$

Part (ii)

Substituting the expressions from Part (i) into the original inequality:

$$(a - 1)(a^{n-1} + a^{n-2} + \dots + a + 1) > n\, a^{\frac{n-1}{2}}(a - 1)$$

Since $a > 1$, it follows that $a - 1 > 0$. We may divide both sides by $(a - 1)$ without flipping the inequality:

$$a^{n-1} + a^{n-2} + \dots + a + 1 > n\, a^{\frac{n-1}{2}}$$

Rearranging the terms in ascending order yields the required equivalent form:

$$1 + a + a^2 + \dots + a^{n-1} > n\, a^{\frac{n-1}{2}} \qquad \square$$

Part (iii) — Method 1: Using AM-GM

Consider the $n$ positive terms $1, a, a^2, \dots, a^{n-1}$. Since $a > 1$ these terms are not all equal, so the AM-GM inequality is strict:

$$\frac{1 + a + a^2 + \dots + a^{n-1}}{n} > \sqrt[n]{1 \cdot a \cdot a^2 \cdots a^{n-1}}$$

Evaluate the product by summing the exponents. The exponent is an arithmetic series:

$$0 + 1 + 2 + \dots + (n-1) = \frac{n(n-1)}{2}$$

So the geometric mean is:

$$\left(a^{\frac{n(n-1)}{2}}\right)^{\!\frac{1}{n}} = a^{\frac{n-1}{2}}$$

Substituting back:

$$\frac{1 + a + a^2 + \dots + a^{n-1}}{n} > a^{\frac{n-1}{2}}$$

Multiplying both sides by $n$ gives the result from Part (ii). Multiplying both sides of that by the positive quantity $(a-1)$ reverses the steps of Part (ii), yielding:

$$a^n - 1 > n\!\left(a^{\frac{n+1}{2}} - a^{\frac{n-1}{2}}\right) \qquad \square$$

Part (iii) — Method 2: Symmetric Pairing (Alternative)

From Part (ii) it suffices to prove $1 + a + a^2 + \dots + a^{n-1} > n\, a^{\frac{n-1}{2}}$. Divide both sides by the positive term $a^{\frac{n-1}{2}}$:

$$a^{-\frac{n-1}{2}} + \dots + a^{-\frac{1}{2}} + a^{\frac{1}{2}} + \dots + a^{\frac{n-1}{2}} > n$$

Pair the first and last terms, second and second-last, and so on. For any $x > 0$ with $x \ne 1$, we have $x + \frac{1}{x} > 2$.

• If $n$ is even: There are $\frac{n}{2}$ pairs, each summing to more than $2$, giving a total strictly greater than $\frac{n}{2} \times 2 = n$.
• If $n$ is odd: There are $\frac{n-1}{2}$ pairs plus the unpaired middle term $a^0 = 1$. The total is strictly greater than $\left(\frac{n-1}{2} \times 2\right) + 1 = n$.

In both cases the sum exceeds $n$. Multiplying back through by $a^{\frac{n-1}{2}}$ and then $(a-1)$ proves the original inequality. $\square$

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Takeaways

• Look for Hidden Geometric Series: Expressions like $x^n - y^n$ or $\frac{x^n - 1}{x - 1}$ frequently appear in Extension 2 inequalities. Recognising that a polynomial can be written as a sum of $n$ terms is often the trigger for applying AM-GM.
• Evaluating the Product of Powers: When applying AM-GM to a sequence of powers, the geometric mean reduces to an arithmetic progression sum in the exponent. Comfort with $\sum_{k=0}^{n-1} k = \frac{n(n-1)}{2}$ is essential.
• Symmetry and Pairing: The pairing method is a beautiful demonstration of Extension 2 thinking. Dividing by the "middle" power exposes symmetric pairs of the form $x^k + x^{-k}$, allowing the simpler $x + \frac{1}{x} > 2$ inequality to replace the full $n$-term AM-GM.

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Further Readings

HSC Induction, HSC Sequences