JC1 H3 Maths Bukit Timah | AM-GM Inequality

JC1 H3 Mathematics Tutor Bukit Timah | AM-GM Inequality and Optimisation. Did you know? Many maximum-and-minimum problems can be solved without differentiation. The trick is to recognise a relationship between the arithmetic mean and the geometric mean—and to check when equality is possible.

For JC1 H2 Mathematics learners hoping to explore H3 Mathematics later, the arithmetic mean–geometric mean (AM–GM) inequality is a natural place to practise proof, assumptions and optimisation in the same question.

Quick answer: what does AM-GM say?

For two positive real numbers a and b, the arithmetic mean is (a+b)/2, and the geometric mean is √(ab). AM–GM states that (a+b)/2 ≥ √(ab). Equality holds precisely when a = b.

That final equality condition is not a decorative extra. If a problem asks for an actual maximum or minimum, a bound is not enough; you must also demonstrate that the bound can be reached under the given constraints.

A short proof you can reconstruct

Because a and b are positive, their square roots are real. The square (√a − √b)² cannot be negative. Expand it: a + b − 2√(ab) ≥ 0.

Rearrange to obtain (a+b)/2 ≥ √(ab). Equality holds exactly when √a = √b, or a = b. This proof is short, but it includes the two things often missed in rushed solutions: why the quantity is nonnegative and when it becomes zero.

Worked example 1: the greatest product with a fixed sum

Question: Two positive numbers x and y satisfy x+y = 12. What is the largest possible value of xy?

AM–GM gives (x+y)/2 ≥ √(xy). Hence 6 ≥ √(xy), which implies xy ≤ 36. Equality requires x = y; together with x+y = 12, that means x = y = 6.

Answer: the maximum product is 36, attained at x = y = 6. We have identified both the bound and the values that achieve it.

Worked example 2: the smallest sum with a fixed product

Question: Positive x and y satisfy xy = 16. Find the least possible value of x+y.

AM–GM gives x+y ≥ 2√(xy) = 2√16 = 8. Equality occurs at x = y, and xy = 16 then gives x = y = 4.

Answer: the minimum sum is 8 at x = y = 4. Did you see the reversal? A fixed sum gave a maximum product; a fixed product gave a minimum sum.

Worked example 3: simplify before choosing calculus

Question: For t > 0, find the minimum of t + 9/t.

Let a = t and b = 9/t. Both are positive, and their product is fixed: ab = 9. Therefore a+b ≥ 2√9 = 6. Equality requires t = 9/t, so t² = 9. Since t > 0, t = 3.

Answer: t + 9/t has minimum 6 at t = 3. If you are using differentiation in H2 Mathematics, you can check the same result by differentiating, but AM–GM reveals why the bound holds without calculus.

Worked example 4: a rectangle with a fixed perimeter

Question: A rectangle has positive side lengths x and y and perimeter 24 units. What is its largest area?

The perimeter equation is 2x+2y = 24, so x+y = 12. The area is A = xy. From the fixed-sum result, A ≤ 36 square units, with equality when x = y = 6.

Therefore, among rectangles with that perimeter, the square has the greatest area. In a word problem, always state both the mathematical value and the physical dimensions.

A tempting mistake: using AM–GM outside its domain

Suppose someone writes AM–GM for a = −1 and b = −4. Their product is positive, but the familiar real AM–GM inequality does not apply because the two inputs are not nonnegative. In fact, (−1−4)/2 = −2.5 is not greater than √4 = 2.

A positive product does not mean the two factors are positive. This is exactly why a proof should state its assumptions before using an inequality.

The equality test that many students overlook

Write the variables and their permitted domain: positive, nonnegative or unrestricted real values.
Identify what the problem keeps constant: a sum, a product or another expression.
Apply the inequality only after choosing nonnegative terms that match its assumptions.
Find the equality condition, such as a=b, and solve it alongside the original constraint.
Return to the requested quantity and explain why the extremum is attained, not merely bounded.

A two-part self-test

Test A: If u > 0, find the smallest possible value of 2u + 8/u. Apply AM–GM to the positive terms 2u and 8/u: the minimum is 2√16 = 8, attained when 2u = 8/u, or u = 2.

Test B: If p and q are positive with p+q = 20, determine the maximum of pq. The maximum is 100 at p = q = 10. Explain why just substituting p=10 into a guessed answer is not a general proof.

Frequently asked questions

Is AM–GM included in H3 Mathematics?

Yes. It appears under the additional inequalities in the 2027 H3 Mathematics 9820 syllabus, together with Cauchy–Schwarz and the triangle inequality.

Should a JC1 student learn H3 inequalities before H2 algebra?

Not at the expense of the foundation. Rearranging expressions, understanding domains and checking equality conditions should already be comfortable. These examples are a way to deepen that understanding.

Does every inequality require AM–GM?

No. Some are more naturally proved by completing the square, factoring a nonnegative expression or applying another inequality. Choosing the right method is part of learning.

Where this fits in the official H3 Mathematics pathway

The Singapore–Cambridge H3 Mathematics 9820 syllabus for 2027 sets out mathematical statements, proof, non-routine problem solving and additional mathematical techniques. It assumes knowledge of H2 Mathematics. See the SEAB 2027 A-Level subject directory for the examination-year reference. These examples are introductory learning exercises, not reproductions of actual examination questions.

A JC1 student can use these ideas to build readiness for a potential H3 pathway, but H3 offering, selection and teaching arrangements depend on the student’s JC. Strengthen the core first with the Bukit Timah H1/H2 Mathematics guide and this H2-to-H3 readiness roadmap.

JC1 H3 Mathematics Tutor Bukit Timah | AM-GM Inequality and Optimisation