AM–GM–HM Inequality: Statement, Proof & Uses

The AM-GM-HM inequality is a single chain that orders the three classical averages: for any set of positive numbers, the arithmetic mean is greater than or equal to the geometric mean, which is greater than or equal to the harmonic mean. In symbols, AM ≥ GM ≥ HM, with equality holding only when all the numbers are identical. For 2 and 8: AM = 5, GM = 4, HM = 3.2, and indeed 5 ≥ 4 ≥ 3.2.

This inequality matters because it turns optimisation problems into one-line arguments. Among all rectangles with a fixed perimeter, which has the largest area? The inequality answers immediately: the square. It proves countless contest inequalities and explains why compound investment returns always trail simple averages.

Key Concepts

• Statement: for positive numbers x_1, …, x_n: (sum)/n ≥ (product)^(1/n) ≥ n/(sum of reciprocals).
• Two-number form: (a + b)/2 ≥ square root of ab ≥ 2ab/(a + b), for a, b > 0.
• Equality condition: equality holds at any step only when all the numbers are equal. If the numbers differ, every inequality is strict.
• Positivity requirement: the chain is guaranteed only for positive (or non-negative, for the AM-GM part) inputs.
• Proof idea for two numbers: from (a - b)^2 ≥ 0: a^2 + b^2 ≥ 2ab, so (a + b)^2 ≥ 4ab, and square roots give (a + b)/2 ≥ square root of ab.

Formulas

• AM-GM (n numbers): (x_1 + … + x_n)/n ≥ (x_1 times … times x_n)^(1/n).
• GM-HM (n numbers): (x_1 times … times x_n)^(1/n) ≥ n/(1/x_1 + … + 1/x_n).
• Full chain: AM ≥ GM ≥ HM, equality iff x_1 = x_2 = … = x_n.
• Assumptions: all x_i > 0 for the full chain; x_i ≥ 0 suffices for AM ≥ GM alone.

Worked Example

First, verify the chain on 2 and 8: AM = (2 + 8)/2 = 5; GM = square root of 16 = 4; HM = (2 times 2 times 8)/(2 + 8) = 32/10 = 3.2. So 5 ≥ 4 ≥ 3.2 holds. Now the optimisation payoff: among rectangles with perimeter 20, let the sides be x and y with x + y = 10. The area is xy, and AM-GM gives square root of xy ≤ (x + y)/2 = 5, so xy ≤ 25, with equality only at x = y = 5: the 5-by-5 square wins with area 25, beating 4 by 6 with area 24.

Common Misconceptions

• Equality holds often: it does not. Equality at any link of the chain demands all numbers be exactly equal; with 2 and 8 every step is a strict inequality.
• It works for negative numbers: it fails. Take a = -4, b = -1: the AM is -2.5 while the GM (square root of 4) is 2, so AM ≥ GM is false. The inequality needs non-negative inputs.
• The GM-HM direction is flipped: some write HM ≥ GM, but the order is AM ≥ GM ≥ HM: the harmonic mean is the smallest.

Real-World Applications

• Optimisation: proving the square maximises area for fixed perimeter, or the cube maximises volume for fixed surface area, without calculus.
• Finance: explaining why a portfolio's compound (geometric) return always trails its average (arithmetic) annual return when returns vary.
• Competitions: the AM-GM step is the single most-used inequality in olympiad algebra problems.

The AM-GM-HM inequality chains the three averages as AM ≥ GM ≥ HM for positive numbers, with equality only when all values match. Its two-number proof needs just (a - b)^2 ≥ 0, yet it settles optimisation questions like the maximum-area rectangle in a line.