Product Inequality with Constraint via AM-GM

This problem is a showcase of the homogenisation technique: substituting a constraint ($a + b + c = 1$) directly into an expression to unlock symmetric structure. The three-part scaffold guides you from a simple algebraic manipulation to a full product inequality, while a slick alternative solution shows how expanding the LHS and invoking standard sub-inequalities can reach the bound in one clean sweep.

Problem Statement

Let $a, b,$ and $c$ be positive real numbers such that $a + b + c = 1$.

i. By writing $\frac{1}{a} + 1$ as a single fraction, show that $\dfrac{1}{a} + 1 = \dfrac{2a+b+c}{a}$.

ii. By applying the AM-GM inequality to the terms $a, a, b,$ and $c$, show that:

$$\frac{1}{a} + 1 \ge \frac{4\sqrt[4]{a^2bc}}{a}$$

iii. Hence, prove that:

$$\left(\frac{1}{a} + 1\right)\!\left(\frac{1}{b} + 1\right)\!\left(\frac{1}{c} + 1\right) \ge 64$$

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Hints

• Part (i): Start by finding a common denominator for $\frac{1}{a} + 1$, giving $\frac{1+a}{a}$. Then use the given condition $a + b + c = 1$ to replace the $1$ in the numerator.
• Part (ii): The numerator from Part (i) is $2a + b + c$. Split the $2a$ into $a + a$ so you have exactly four positive terms, then apply AM-GM for $n = 4$.
• Part (iii): Because the inequality is symmetric in $a, b, c$, establish analogous inequalities for the factors containing $b$ and $c$. Multiply the three inequalities together and simplify the resulting powers.

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Solutions

Part (i)

$$\frac{1}{a} + 1 = \frac{1 + a}{a}$$

Since $a + b + c = 1$, substitute into the numerator:

$$\frac{1 + a}{a} = \frac{(a+b+c) + a}{a} = \frac{2a + b + c}{a} \qquad \square$$

Part (ii)

Consider the four positive numbers $a, a, b, c$. By the AM-GM inequality for four terms:

$$\frac{a + a + b + c}{4} \ge \sqrt[4]{a \cdot a \cdot b \cdot c}$$ $$2a + b + c \ge 4\sqrt[4]{a^2bc}$$

Dividing both sides by the strictly positive quantity $a$:

$$\frac{2a + b + c}{a} \ge \frac{4\sqrt[4]{a^2bc}}{a}$$

Substituting the result from Part (i) yields:

$$\frac{1}{a} + 1 \ge \frac{4\sqrt[4]{a^2bc}}{a} \qquad \square$$

Part (iii) — Method 1: AM-GM on Individual Factors

By symmetry, applying the identical argument to the other two factors gives:

$$\frac{1}{b} + 1 \ge \frac{4\sqrt[4]{ab^2c}}{b}, \qquad \frac{1}{c} + 1 \ge \frac{4\sqrt[4]{abc^2}}{c}$$

Since all factors are positive, multiply the three inequalities:

$$\left(\frac{1}{a}+1\right)\!\left(\frac{1}{b}+1\right)\!\left(\frac{1}{c}+1\right) \ge \frac{4\sqrt[4]{a^2bc}}{a} \cdot \frac{4\sqrt[4]{ab^2c}}{b} \cdot \frac{4\sqrt[4]{abc^2}}{c}$$

Combine the numerator using the product of fourth roots:

$$\ge \frac{64\,\sqrt[4]{(a^2bc)(ab^2c)(abc^2)}}{abc} = \frac{64\,\sqrt[4]{a^4b^4c^4}}{abc} = \frac{64\,abc}{abc} = 64 \qquad \square$$

Part (iii) — Method 2: Direct Expansion (Alternative)

Expand the left-hand side:

$$LHS = 1 + \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + \left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right) + \frac{1}{abc}$$

Notice that $\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca} = \dfrac{c + a + b}{abc} = \dfrac{1}{abc}$ (using $a+b+c=1$). So:

$$LHS = 1 + \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + \frac{2}{abc}$$

Now bound each piece using two standard AM-GM results:

1. Sum of reciprocals: By AM-HM (or AM-GM applied to $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$):

$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{9}{a+b+c} = 9$$

2. Product bound: AM-GM on $a + b + c = 1$ gives $\dfrac{1}{3} \ge \sqrt[3]{abc}$, so $abc \le \dfrac{1}{27}$, hence $\dfrac{1}{abc} \ge 27$.

Substituting these lower bounds:

$$LHS \ge 1 + 9 + 2(27) = 1 + 9 + 54 = 64 \qquad \square$$

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Takeaways

• Homogenisation: Substituting a constraint like $a + b + c = 1$ directly into an expression is a powerful technique. It converts constants into symmetric variable expressions that are immediately amenable to AM-GM.
• Splitting Terms: Part (ii) requires recognising that $2a$ must be written as $a + a$ to generate a four-term AM-GM. This is deliberate: the symmetry of the split ensures the fourth roots cancel cleanly when the three inequalities are multiplied.
• The Power of Expansion: The alternative solution highlights the value of recognising standard sub-inequalities ($\sum \frac{1}{x} \ge \frac{n^2}{\sum x}$). Expanding a product in an inequality problem often reveals independent components that can each be bounded individually, sidestepping the scaffold entirely.

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Further Readings

HSC Induction, HSC Sequences