Geometric Mean Inequalities
On this page
Direct answer
a + b ≥ 2√(ab) is the most reused line in the entire paper: for positive reals, the arithmetic mean never falls below the geometric mean, with equality only when the numbers coincide. The full chain for three numbers is (a + b + c)/3 ≥ (abc)^(1/3) ≥ 3/(1/a + 1/b + 1/c), every equality collapsing to a = b = c. Two-way use drives the exam questions: a fixed sum maximises the product at equal splitting, and a fixed product minimises the sum at the same point. The weighted version Σλᵢaᵢ ≥ Πaᵢ^(λᵢ) (λᵢ > 0, Σλᵢ = 1) handles unequal weights.
What you must remember
- Two-variable core: a + b ≥ 2√(ab), equality iff a = b; consequence x + k/x ≥ 2√k for x > 0, minimum at x = √k.
- Three variables: a + b + c ≥ 3(abc)^(1/3); fixed sum s forces abc ≤ (s/3)³, attained at a = b = c = s/3.
- Cubic identity route: a³ + b³ + c³ − 3abc = ½(a + b + c)[(a − b)² + (b − c)² + (c − a)²] ≥ 0 for non-negative a, b, c — equality iff a = b = c.
- Weighted AM-GM: λ₁a₁ + ... + λ_k a_k ≥ a₁^λ₁···a_k^λ_k when λᵢ > 0 sum to 1; equality again demands all aᵢ equal.
- Cyclic sums: for positive a, b, c, the sum a/b + b/c + c/a ≥ 3, since the three summands multiply to 1.
- Reciprocal mileage: AM ≥ HM gives (a + b + c)(1/a + 1/b + 1/c) ≥ 9 — the engine behind most fraction-sum minima.
- Positivity guard: every AM-GM form needs strictly positive entries; applying it to quantities that can be zero or negative voids the inequality and its equality condition.
Two maximisation problems
Problem one: with a + b + c = 6 and a, b, c > 0, maximise abc. AM-GM gives 6/3 ≥ (abc)^(1/3), so abc ≤ 8, equality at a = b = c = 2. The equality check is not decoration — it converts an inequality into an attained maximum, and quoting it is what separates a complete solution from a bound. Problem two: minimise f = a/b + b/c + c/a over positive reals. The three summands are positive with product (a/b)(b/c)(c/a) = 1, so f ≥ 3·1^(1/3) = 3, with equality when a/b = b/c = c/a = 1, i.e. a = b = c. Minimum 3, attained.
Now the reflex-breaker: minimise x + 3/x for x > 0. AM-GM gives x·(3/x) = 3, so x + 3/x ≥ 2√3, equality at x = √3 — not at x = 1. The number under the root comes from the product of the two summands, and quoting "minimum 2" (the x + 1/x reflex) would be wrong here. The same adjustment governs every weighted variant: x + 12/x bottoms out at 4√3, x + k/x at 2√k.
When AM-GM fails
Main uses these as single-correct or numerical questions with quotable answers — 8, 3, 2√3 — and expects the equality condition in the working. Advanced breaks naive applications in two characteristic ways. First, coupled constraints: when the variables satisfy an equation, separate AM-GM bounds can each demand equality at incompatible points, so the true extremum is larger and a different method (substitution, calculus, Cauchy) is required — always verify the equality point actually satisfies every constraint before announcing the minimum. Second, sign violations: expressions containing quantities like a + b that might be negative, or terms that vanish, cannot be fed to AM-GM at all; a bound derived on a forbidden domain is not a bound. The examiner's tell is an option set where two answers differ by exactly the gap between "equality achievable" and "equality impossible".
Frequently asked questions
What is the minimum of x + 4/x for x > 0?
2√4 = 4, attained at x = 2 — AM-GM on summands whose product is the constant 4.
When does AM equal GM?
Exactly when all the numbers involved are equal; any unequal pair makes the inequality strict.
How does AM-GM maximise a product under a fixed sum?
If a + b + c = s, then abc ≤ (s/3)³ with the maximum at a = b = c = s/3; the equality condition makes the bound attainable.
Can AM-GM be applied to negative numbers?
No — the standard forms require strictly positive entries, and with mixed signs neither the direction of the inequality nor the equality condition survives.
What does AM ≥ HM give for fraction sums?
(a + b + c)(1/a + 1/b + 1/c) ≥ 9 for positive a, b, c — the workhorse behind reciprocal-sum minima.