The AM-GM Inequality

Why This Matters

The arithmetic-geometric mean inequality is the most-used inequality in olympiad mathematics. Its statement:

For non-negative reals $a_1, a_2, \ldots, a_n$ , $\frac{a_1 + a_2 + \cdots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \cdots a_n},$ with equality if and only if $a_1 = a_2 = \cdots = a_n$ .

Every olympiad student should know AM-GM cold: the statement, the equality condition, at least one proof, and recognition cues. The technique applies whenever a problem mixes additive and multiplicative structure, which is most non-trivial inequalities you will encounter.

Recognize AM-GM when:

You have a sum bounded above and want a product bound below (or the reverse).
The expression has an obvious "balance point" where all variables are equal (equality conditions are a hint).
A problem asks for the maximum of a product subject to a fixed sum (or minimum of a sum subject to a fixed product); AM-GM gives the bound and the equality condition gives the optimum.
The problem has homogeneous structure: scale all variables by $\lambda$ and the inequality scales predictably.

The two-variable case $\frac{a + b}{2} \ge \sqrt{a b}$ is elementary: it rearranges to $(\sqrt{a} - \sqrt{b})^2 \ge 0$ . The general case for $n$ variables is more delicate. Three common proofs cover most situations.

The Inequality

Theorem

AM-GM Inequality

Statement

$\frac{a_1 + a_2 + \cdots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \cdots a_n}$ , with equality iff $a_1 = a_2 = \cdots = a_n$ .

Intuition

"Spreading" mass equally maximizes a product subject to a fixed sum. If two of the $a_i$ differ, replace them with their average; the sum is unchanged but the product strictly increases (since $a b < \left(\frac{a + b}{2}\right)^2$ when $a \ne b$ ). Iterating this smoothing converges to the all-equal configuration, which has product $\bar{a}^n$ where $\bar{a}$ is the common value, equal to the AM. So the AM is the supremum of the GM under fixed sum.

Proof Sketch

Cauchy's forward-backward induction (1821). The proof uses reverse induction, a rare and elegant technique.

Forward step (powers of 2). The case $n = 2$ is elementary: $\frac{a + b}{2} \ge \sqrt{a b} \Leftrightarrow (\sqrt{a} - \sqrt{b})^2 \ge 0$ . From $n$ to $2n$ : applying the case $n = 2$ to the two arithmetic means $\frac{a_1 + \cdots + a_n}{n}$ and $\frac{a_{n+1} + \cdots + a_{2n}}{n}$ gives $\frac{a_1 + \cdots + a_{2n}}{2n} \ge \sqrt{\frac{(a_1 + \cdots + a_n)(a_{n+1} + \cdots + a_{2n})}{n^2}} \ge \sqrt[2n]{a_1 \cdots a_{2n}}.$ By induction the inequality holds for every $n$ that is a power of 2.

Backward step. Suppose AM-GM holds for $n + 1$ variables; we show it holds for $n$ . Given $a_1, \ldots, a_n$ , set $a_{n+1} = \frac{a_1 + \cdots + a_n}{n}$ (the AM of the others). Apply AM-GM at $n + 1$ to $a_1, \ldots, a_{n+1}$ : $\frac{a_1 + \cdots + a_{n+1}}{n+1} \ge \sqrt[n+1]{a_1 \cdots a_{n+1}}.$ The left side equals $a_{n+1}$ (by construction). Solving for $a_1 \cdots a_n$ gives $a_{n+1}^n \ge a_1 \cdots a_n$ , i.e., the AM-GM inequality at $n$ .

Combining: forward induction covers all powers of 2, and backward induction descends to fill in the gaps. Every $n$ is covered.

Equality: each step preserves equality iff all the values are equal. Trace through to verify.

Why It Matters

Cauchy's forward-backward induction is one of the most beautiful proof structures in elementary mathematics. The proof exploits a structural feature of $\mathbb{N}$ (every integer is at most a power of 2 and at least 1) to extend a binary inequality to all $n$ . The shape recurs in some generalized inequality proofs and in measure-theoretic arguments where you "double up" then descend.

Other proofs of AM-GM:

Smoothing. Repeatedly replace pairs with their averages; show the product strictly increases until all equal.
Jensen on $\log$ . Apply Jensen's inequality to the concave function $\log x$ : $\frac{\sum \log a_i}{n} \le \log \frac{\sum a_i}{n}$ , exponentiate.
Lagrange multipliers (for the optimization-flavored version). Maximize $\prod a_i$ subject to $\sum a_i = S$ ; the critical point has all $a_i$ equal.

Failure Mode

The hypothesis " $a_i \ge 0$ " is essential. With negative values the geometric mean is not well-defined for even $n$ (negative under a $\sqrt[2]{}$ ), and for odd $n$ the inequality can fail. Counterexample: $a_1 = -2$ , $a_2 = -2$ , $a_3 = -2$ . Then AM $= -2$ , GM $= -2$ , equality holds; but try $a_1 = -2$ , $a_2 = 1$ , $a_3 = 1$ : AM $= 0$ , GM $= \sqrt[3]{-2}$ , and the comparison is sign-dependent.

In practice, every clean olympiad use of AM-GM either has positive variables or has a substitution that makes them positive. If you find yourself negating to apply AM-GM to a "negative" quantity, restructure first.

report a correction →

Worked Example: Maximize a Product

Example

Maximize a product with fixed sum

Problem. Find the maximum of $x y z$ given $x + y + z = 12$ with $x, y, z > 0$ .

Solution. AM-GM with $n = 3$ : $\frac{x + y + z}{3} \ge \sqrt[3]{x y z} \quad \Rightarrow \quad 4 \ge \sqrt[3]{x y z} \quad \Rightarrow \quad x y z \le 64.$ Equality holds iff $x = y = z = 4$ . So the maximum is 64, attained at $(4, 4, 4)$ .

This is the canonical AM-GM application: a sum is fixed, you want a product bound. The bound is automatic; the equality condition tells you the optimum is at the symmetric point.

Worked Example: A Three-Term Inequality

Example

A three-term cyclic inequality

Problem. Show that for positive reals $a, b, c$ , $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \ge a + b + c.$

Solution. Apply AM-GM to two terms at a time: $\frac{a^2}{b} + b \ge 2 \sqrt{\frac{a^2}{b} \cdot b} = 2 a, \quad \frac{b^2}{c} + c \ge 2 b, \quad \frac{c^2}{a} + a \ge 2 c.$

Summing the three: $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} + (a + b + c) \ge 2(a + b + c)$ .

Subtracting $a + b + c$ from both sides: $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \ge a + b + c$ . ✓

Equality requires $\frac{a^2}{b} = b$ , $\frac{b^2}{c} = c$ , $\frac{c^2}{a} = a$ , i.e., $a = b = c$ .

This is the pairing technique: introduce a clever companion term, apply two-variable AM-GM, sum.

The Recognition Process

When facing an inequality, ask:

Are the variables non-negative? AM-GM needs this. Often the problem hypothesis ensures it; sometimes you need a substitution.
Is the expression homogeneous? Homogeneous inequalities normalize cleanly (set $\sum a_i = 1$ or $\prod a_i = 1$ ), which often turns AM-GM into a one-line argument.
What is the equality case? A symmetric expression typically has equality at $a_1 = \cdots = a_n$ . If so, AM-GM is a strong candidate. If equality is at a non-symmetric point, AM-GM may not be the right tool.
What pairs cleanly? Sometimes you apply AM-GM to a subset of the terms, with a clever "companion" added to make the GM simplify.
Could weighted AM-GM help? $\sum w_i a_i \ge \prod a_i^{w_i}$ for $\sum w_i = 1$ . Use this when the equality case is asymmetric.

Common Mistakes

Watch Out

Forgetting positivity

AM-GM requires $a_i \ge 0$ . If the problem allows negative values, you cannot apply AM-GM directly. Often a substitution ( $b_i = a_i + C$ or $b_i = e^{a_i}$ ) makes everything positive; sometimes the positivity hypothesis is hidden in the problem statement and requires careful reading.

Watch Out

Equality conditions

The equality condition $a_1 = a_2 = \cdots = a_n$ is every $a_i$ equal, not just two of them. When chaining two applications of AM-GM, the joint equality requires all relevant variables equal across both applications. Many competition writeups lose marks here.

Watch Out

Weighted vs unweighted

The unweighted form $\frac{\sum a_i}{n} \ge \sqrt[n]{\prod a_i}$ has equal weight $1/n$ on each term. The weighted form $\sum w_i a_i \ge \prod a_i^{w_i}$ for $\sum w_i = 1$ , $w_i \ge 0$ allows different weights. Confusing the two leads to wrong bounds. If your problem has natural asymmetric weights (e.g., multiple copies of a variable), weighted AM-GM is the correct form.

Watch Out

Direction of the inequality

AM ≥ GM. Beginners sometimes write GM ≥ AM by reflex, especially when applying the two-variable case to deduce $\sqrt{a b} \ge \frac{a + b}{2}$ . The inequality is wrong in that direction. The geometric mean of distinct values is smaller than their arithmetic mean.

Watch Out

Not all symmetric inequalities yield to AM-GM

A symmetric inequality with equality at $a = b = c$ is a hint that AM-GM might work, but it is not a guarantee. Some symmetric inequalities require Schur, SOS, Jensen, or rearrangement. AM-GM is the right tool when the structure mixes sums and products of the variables, not just symmetric combinations.

Exercises

ExerciseCore

Problem

Show that for all positive reals $a$ and $b$ , $\frac{a}{b} + \frac{b}{a} \ge 2$ , with equality iff $a = b$ .

ExerciseCore

Problem

For positive reals $a, b, c$ with $abc = 1$ , show that $a + b + c \ge 3$ .

ExerciseAdvanced

Problem

Show that for positive reals $a, b, c$ with $a + b + c = 3$ , $\frac{a}{1 + b^2} + \frac{b}{1 + c^2} + \frac{c}{1 + a^2} \ge \frac{3}{2}.$

Cross-Network Links

ProofsPath: cauchy-schwarz-inequality is the bilinear cousin; power-mean-inequality is the family AM-GM lives in (AM and GM are the $r = 1$ and $r = 0$ power means); jensen-for-convex-functions generalizes to any concave function (AM-GM is Jensen on $\log$ ); rearrangement-inequality handles cases where AM-GM does not.
TheoremPath: common-probability-distributions uses AM-GM in the proof of entropy bounds and concentration inequalities; maximum-likelihood-estimation uses AM-GM in some asymptotic-efficiency arguments.

References

See structured references block. Primary entry points: Hardy-Littlewood-Polya Inequalities Ch 2 for the canonical treatment with multiple proofs; Steele Cauchy-Schwarz Master Class Ch 2 for olympiad-flavored applications; Engel Problem-Solving Strategies Ch 7 for the contest perspective.