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Many JEE algebra questions ask for the least or greatest value of an expression. Calculus works, but it is slow and the algebra is easy to get wrong. The AM-GM inequality often gives the answer in two lines. It also gives wrong answers in two lines when it is used carelessly. This post shows when it applies, works through six examples, and lists the traps.
What does the AM-GM inequality say?
For positive numbers a and b, the arithmetic mean is (a + b)/2 and the geometric mean is √(ab). NCERT's Class 11 chapter on Sequences and Series proves that A − G = (√a − √b)²/2, which is never negative. So A ≥ G, and A = G only when a = b.
The same holds for more positive numbers. For three positive numbers, (a + b + c)/3 ≥ ∛(abc), again with equality only when all three are equal. NCERT proves the two-number case; the general version is standard and widely used in JEE problems.
The official JEE (Main) 2026 syllabus lists "Relation between A.M and G.M." in its sequence and series unit. Check the syllabus for your own exam year. We cover the rest of that unit in our post on sequences and series for JEE.
When can you use it?
Check three things before you write a single line.
Positive terms. Every term in the mean must be positive. For negative x, AM-GM simply does not apply.
Something fixed. If the product of the terms is constant, AM-GM gives a minimum for their sum. If the sum is constant, it gives a maximum for their product.
Equality must be possible. The bound is only the answer if the terms can actually be equal for an allowed value of the variable.
Worked examples
1. Least value of 4x + 9/x for x > 0. Both terms are positive and their product is 36, a constant. So (4x + 9/x)/2 ≥ √36 = 6, which gives 4x + 9/x ≥ 12. Equality needs 4x = 9/x, so x² = 9/4 and x = 3/2. That is allowed, so the least value is 12. Check: 4(1.5) + 9/1.5 = 6 + 6 = 12.
2. A fixed perimeter. A rectangle has perimeter 40 cm, so its sides satisfy a + b = 20. Then √(ab) ≤ (a + b)/2 = 10, so ab ≤ 100. The largest area is 100 cm², reached when a = b = 10, a square.
3. Three numbers with a fixed sum. If a, b and c are positive and a + b + c = 12, then ∛(abc) ≤ 12/3 = 4, so abc ≤ 64. Equality at a = b = c = 4.
4. Splitting a term. Find the least value of x + 2/x² for x > 0. The product x · (2/x²) is not constant, so split x into two equal halves: x/2 + x/2 + 2/x². Now the product of the three terms is (x/2)(x/2)(2/x²) = 1/2, a constant. AM-GM gives
x/2 + x/2 + 2/x² ≥ 3 ∛(1/2)
Equality needs x/2 = 2/x², so x³ = 4. The least value is 3/∛2, about 2.38. Calculus agrees: the derivative 1 − 4/x³ is zero at x³ = 4, and at x = ∛4 ≈ 1.587 the expression is about 1.587 + 0.794 = 2.381.
5. Proving an inequality. Show that (a + b)(b + c)(c + a) ≥ 8abc for positive a, b, c. Apply AM-GM to each bracket: a + b ≥ 2√(ab), b + c ≥ 2√(bc), c + a ≥ 2√(ca). Multiply the three: the right side is 8√(a²b²c²) = 8abc. Multiplying is allowed here because every side is positive.
6. A quick check of a classic. For positive a and b, (a + b)(1/a + 1/b) ≥ 4. Each bracket is at least twice a square root, and the roots multiply to √(ab) · √(1/(ab)) = 1, so the product is at least 2 × 2 = 4.
Which traps cost the most marks?
Equality outside the domain. Find the least value of x + 1/x for x ≥ 2. AM-GM says x + 1/x ≥ 2, but equality needs x = 1, which is not allowed. For x ≥ 2 the expression increases, so the least value is 2 + 1/2 = 2.5, not 2.
The trigonometric version of the same trap. For sin²θ + 4/sin²θ, AM-GM gives at least 4, with equality when sin²θ = 2. That is impossible. Since sin²θ is at most 1, and t + 4/t decreases for 0 < t < 2, the least value is at sin²θ = 1: 1 + 4 = 5.
Negative terms. For x < 0, x + 1/x is at most −2, not at least 2. Check signs first.
Nothing fixed. If neither the sum nor the product is constant, the inequality gives a bound that depends on x, which is not an answer. Try splitting terms, as in example 4, or switch to calculus.
A weekly AM-GM drill
Sort before solving (10 minutes). Take eight min/max problems and, for each, write whether AM-GM applies and which quantity is fixed. Do not solve yet.
Solve with the equality line (20 minutes). Solve the ones that qualify, and write the equality case as the last line of every solution.
Check one by calculus (5 minutes). Pick one problem and confirm it by differentiation. This catches equality traps.
Mix it in. Put AM-GM problems inside sets with quadratics and calculus maxima so you must choose the method. In a study of 126 seventh-grade maths students, Rohrer, Dedrick and Stershic (2015) found that interleaved practice produced higher scores than practising one type at a time, on tests one day and 30 days later.
When a problem will not split into terms with a fixed product and you cannot see why, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can compare its first step with yours.
Start today: find the least value of 9x + 1/x for x > 0, and write the equality case before you check the answer.
