Geometric mean and its applications — AM-GM inequality

Question

(a) Find the geometric mean of 4 and 16. (b) If three positive numbers are in GP with common ratio $r$ , show that their AM $\geq$ GM. (c) Using AM-GM inequality, find the minimum value of $x + \dfrac{9}{x}$ for $x > 0$ .

(CBSE 11 + JEE Main pattern)

Solution — Step by Step

The geometric mean of two positive numbers $a$ and $b$ :

GM = \sqrt{ab} = \sqrt{4 \times 16} = \sqrt{64} = \mathbf{8}

Notice: $4, 8, 16$ form a GP with ratio 2. The GM always sits “between” the two numbers in a multiplicative sense.

For positive $a$ and $b$ :

AM = \frac{a + b}{2}, \quad GM = \sqrt{ab}

We need to show $\dfrac{a+b}{2} \geq \sqrt{ab}$ .

\frac{a+b}{2} - \sqrt{ab} = \frac{a + b - 2\sqrt{ab}}{2} = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \geq 0

Since a square is always $\geq 0$ , we get $AM \geq GM$ , with equality when $a = b$ .

Apply AM-GM to the two positive terms $x$ and $\dfrac{9}{x}$ :

\frac{x + \frac{9}{x}}{2} \geq \sqrt{x \cdot \frac{9}{x}} = \sqrt{9} = 3

x + \frac{9}{x} \geq 6

Minimum value is $\mathbf{6}$ , achieved when $x = \dfrac{9}{x} \implies x^2 = 9 \implies x = 3$ .

flowchart TD
    A["AM-GM Inequality"] --> B["For positive a, b: AM ≥ GM"]
    B --> C["(a+b)/2 ≥ √(ab)"]
    C --> D["Equality when a = b"]
    D --> E["Application: Find min/max"]
    E --> F["Split expression into parts"]
    F --> G["Apply AM ≥ GM to the parts"]
    G --> H["Product of parts must be constant"]
    H --> I["Minimum of sum = 2√(product)"]

Why This Works

The AM-GM inequality is fundamentally about the relationship between addition and multiplication. The arithmetic mean weights things equally by sum; the geometric mean weights them by product. For positive numbers, the “additive average” is always at least as large as the “multiplicative average.”

The optimization trick works because if the product of two terms is constant (here $x \cdot \dfrac{9}{x} = 9$ is fixed regardless of $x$ ), then their sum is minimised when both terms are equal. This is a powerful shortcut — no calculus needed!

Alternative Method — Calculus Verification

Let $f(x) = x + \dfrac{9}{x}$ . Setting $f'(x) = 0$ :

f'(x) = 1 - \frac{9}{x^2} = 0 \implies x^2 = 9 \implies x = 3

f''(3) = \frac{18}{27} > 0 \text{ (minimum confirmed)}

$f(3) = 3 + 3 = 6$ ✓

AM-GM is a JEE favourite for optimization without calculus. The trick: split the expression so that the PRODUCT of the parts is a constant. For $x + \dfrac{9}{x}$ , the product is 9 (constant). For $x^2 + \dfrac{4}{x}$ , split as $\dfrac{x^2}{2} + \dfrac{x^2}{2} + \dfrac{4}{x}$ (three terms) and apply AM-GM to three terms: $AM \geq \sqrt[3]{product}$ .

Common Mistake

Students apply AM-GM to expressions where the terms can be negative or zero. AM-GM only works for POSITIVE numbers. If a problem says $x$ can be any real number and asks to minimise $x + \dfrac{9}{x}$ , AM-GM gives minimum 6 only for $x > 0$ . For $x < 0$ , the minimum is actually $-6$ (at $x = -3$ ). Always check the domain before applying AM-GM.

Question

Solution — Step by Step

Why This Works

Alternative Method — Calculus Verification

Common Mistake

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