AM–GM, Cauchy–Schwarz and the triangle inequality

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An inequality proof shows that one expression can never be smaller than another, for every allowed value of the variables. Two classical inequalities do most of the work at H3 level. The AM–GM inequality compares a sum with a product, and the Cauchy–Schwarz inequality compares a sum of products with sums of squares. This lesson proves both from the fact that a square is never negative, uses them to find least and greatest values, and shows the two checks that examiners look for: that every step is valid for the stated variables, and that the equality case can actually occur.

AM–GM from a square

For real a,b ≥ 0, (square root of a - square root of b)² ≥ 0 expands to a + b-2 square root of ab ≥ 0. Rearranging gives the two-variable AM–GM inequality:

(a + b)/2 ≥ square root of ab (a,b ≥ 0),

with equality exactly when square root of a = square root of b, that is a = b. The arithmetic mean is at least the geometric mean. More generally, for non-negative a₁,…,aₙ,

(a₁ + a₂ + … + aₙ)/n ≥ nth root of (a₁a₂… aₙ),

with equality exactly when all the aᵢ are equal.

AM–GM turns a fixed product into a lower bound for a sum. For x > 0,

x + 4/x ≥ 2 square root of (x · 4/x) = 4,

with equality when x = 4/x, that is x = 2. So the least value of x + 4/x is 4. A bound is a least value only if its equality case is allowed.

Find the least value of x² + 2/x for x > 0, without calculus.

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Choose terms with a constant product
Method
Write 2/x = 1/x + 1/x and apply AM–GM to the three terms x², 1/x, 1/x.
Reason
Applying two-term AM–GM to x² and 2/x gives 2 square root of 2x, which still depends on x. Splitting makes the product x² · 1/x · 1/x = 1 constant.
Working
Answer
x² + 1/x + 1/x ≥ 3 cube root of (x² · 1/x · 1/x); = 3.
Check that equality can occur
Working
Answer
Equality needs x² = 1/x, so x = 1, which is allowed. The least value is 3, attained at x = 1.

Quick check

Calculus agrees: the derivative 2x-2/x² is zero only at x = 1. The art in AM–GM is choosing the terms so that their product is constant and they can all be equal at once.

A student is asked for the least value of x + 1/x for x ≥ 2. The student writes x + 1/x ≥ 2 square root of (x · 1/x) = 2 and answers 2. Explain the error and find the least value.

Write your answer, then compare it with the solution

The error and the least value

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Test the equality case
Method
Equality in AM–GM needs x = 1/x, that is x = 1.
Reason
x = 1 is not in the domain x ≥ 2, so the value 2 is a true lower bound but is never attained.
Working
Answer
For x ≥ 2,
(x + 1/x)-5/2 = (2x²-5x + 2)/(2x) = ((2x-1)(x-2))/(2x) ≥ 0,
since 2x-1 > 0, x-2 ≥ 0 and x > 0. Equality holds at x = 2, so the least value is 5/2.

Common mistake

The inequality x + 1/x ≥ 2 is correct; the conclusion “the least value is 2” is not. A least value needs both a bound and an allowed input that reaches it.

Prove that for all positive real numbers a, b, c,

(a + b)(b + c)(c + a) ≥ 8abc,

and state when equality holds. [4]

Write your answer, then compare it with the solution

Your proof and the equality case

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Bound each factor
Method
By AM–GM, a + b ≥ 2 square root of ab, b + c ≥ 2 square root of bc, c + a ≥ 2 square root of ca.
Reason
Each factor is a sum of two positive numbers.
Working
Answer
All six quantities are positive, so the inequalities may be multiplied:
(a + b)(b + c)(c + a) ≥ 8 square root of (ab · bc · ca) = 8abc.
Equality
Method
Equality needs equality in all three bounds.
Reason
A product of strict and non-strict inequalities between positives is strict.
Working
Answer
So equality holds exactly when a = b, b = c and c = a, that is a = b = c. blacksquare

Self-mark with the mark scheme

Tick each mark point that your answer includes.

Tick a point only if your written proof contains it. This self-check does not record a mark.

Mark points

1 mark: Applies AM–GM correctly to each of the three factors.
1 mark: Justifies multiplying the inequalities because every quantity is positive.
1 mark: Simplifies the product of the square roots to abc to reach 8abc.
1 mark: States that equality holds exactly when a = b = c, with a reason.

Common mistake

Multiplying inequalities is only valid when both sides are nonnegative: -3 ≤ 1 and -3 ≤ 1 do not give 9 ≤ 1. Expanding gives a²b + a²c + ab² + b²c + ac² + bc²-6abc ≥ 0, which is also provable but hides the structure.

Cauchy–Schwarz from a quadratic

For real a₁,…,aₙ and b₁,…,bₙ,

(∑ᵢ₌₁ⁿ aᵢbᵢ)² ≤ (∑ᵢ₌₁ⁿ aᵢ²)(∑ᵢ₌₁ⁿ bᵢ²).

Proof. If every aᵢ = 0, both sides are 0. Otherwise consider

q(t) = ∑ᵢ₌₁ⁿ(aᵢt + bᵢ)² = (∑ aᵢ²)t² + 2(∑ aᵢbᵢ)t + ∑ bᵢ².

As a sum of squares, q(t) ≥ 0 for every real t, and its leading coefficient is positive, so its discriminant is at most 0:

4(∑ aᵢbᵢ)²-4(∑ aᵢ²)(∑ bᵢ²) ≤ 0,

which is the inequality. Equality holds exactly when q has a real root t, so that bᵢ = -taᵢ for every i: the two lists are proportional.

In vector form this is |a · b| ≤ |a||b|. The triangle inequality follows:

|a + b|² = |a|² + 2 a · b + |b|² ≤ (|a| + |b|)².

For real numbers, |x + y| ≤ |x| + |y|, with equality exactly when xy ≥ 0.

Real numbers x, y, z satisfy x² + y² + z² = 1. Use Cauchy–Schwarz to find the greatest value of x + 2y + 2z and the point where it occurs.

Your inequality, the greatest value and where it occurs

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With (1,2,2) and (x,y,z): (x + 2y + 2z)² ≤ (1 + 4 + 4)(x² + y² + z²) = 9, so x + 2y + 2z ≤ 3. Equality needs (x,y,z) proportional to (1,2,2) with positive constant and length 1: (x,y,z) = (1/3,2/3,2/3), which gives 1/3 + 4/3 + 4/3 = 3.

Common mistakes

Using AM–GM with a negative term. The inequality needs every term to be non-negative.
Multiplying inequalities whose sides may be negative.
Stating a least or greatest value without checking that the equality case satisfies every condition in the question.
Applying AM–GM to terms whose product still depends on the variable, which gives a bound that is not a constant.
In Cauchy–Schwarz, forgetting that equality needs the two lists to be proportional, and with the right sign for a greatest value.

Before you move on

Find the least value of 4x + 9/x for x > 0.
Positive real numbers x, y, z satisfy xyz = 8. Find the least value of x + y + z.

1. Least value of 4x + 9/x

2. Least value of x + y + z

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4x + 9/x ≥ 2 square root of (4x · 9/x) = 12, with equality when 4x = 9/x, that is x = 3/2. The least value is 12.
x + y + z ≥ 3 cube root of xyz = 3 cube root of 8 = 6, with equality when x = y = z = 2, which satisfies xyz = 8. The least value is 6.

Syllabus and review details

GCE A-Level H3 Mathematics 2027 · 2027

Last reviewed: 2026-11-06

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Guided steps

Reading settings

AM–GM from a square

Split a term to fix the product

Choose terms with a constant product

Check that equality can occur

A bound that is never reached

Test the equality case

A product of three sums

Bound each factor

Equality

Cauchy–Schwarz from a quadratic

A greatest value on a sphere

Common mistakes

Before you move on

Two least values

Lesson done

AM–GM from a square

Split a term to fix the product

A bound that is never reached

A product of three sums

Cauchy–Schwarz from a quadratic

A greatest value on a sphere

Common mistakes

Before you move on

Two least values

Questions