AM–GM Without Memorising: Recognising the Inequality in a Problem
You do not need to recall every named inequality. You need to recognise the shape "a sum is minimised when the parts are equal" and know when the equality case can be attained.
Olympiad Desk · Core Learning
What you will take away
- For positive a, b: (a + b)/2 ≥ √(ab), with equality exactly when a = b
- Chase equality: if a = b is not achievable, the bound is not tight
- Add a clever constant or split terms to make the product constant
- Sum and product relationships are the giveaway that AM–GM applies
Inequality problems are pattern recognition. If the expression mixes a sum with a product, or a quantity with its reciprocal, AM–GM is usually the first thing to try — provided you check equality can actually happen.
The equality check
After applying the inequality, ask whether the equality condition is compatible with the constraints. If the problem forces a = 3 while equality demands a = b, the bound you produced is not sharp and you need a different split — a warning sign that carries a lot of marks.
Splitting terms
When a product is not constant, split one term into several equal parts. Replacing x by x/2 + x/2 lets you tune the product so that everything cancels, which turns a stubborn inequality into a two-line computation.
Worked problems
- 1
For x > 0, find the minimum value of x + 4/x.
Show hint
The product of the two terms is constant here — that is exactly the AM–GM setup.
Show solution
By AM–GM, x + 4/x ≥ 2√(x · 4/x) = 2√4 = 4. Equality requires x = 4/x, i.e. x = 2, which is allowed since x > 0.
Answer: The minimum is 4, attained at x = 2.
- 2
Prove that for positive a and b, a/b + b/a ≥ 2.
Show hint
Multiply by the positive quantity ab, or apply AM–GM directly to the two terms.
Show solution
Apply AM–GM to the positive terms a/b and b/a: their product is 1, so their sum is at least 2√1 = 2, with equality when a = b.
Answer: a/b + b/a ≥ 2, with equality iff a = b.
