AM–GM Inequality and Harmonic Mean: Examples and Practice

Before this lesson: Sum of squares, Arithmetic and exact calculation

Your goal: Choose the terms in a mean inequality and check attainable equality.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Names you may know: AM-GM inequality; arithmetic mean–geometric mean inequality; AM-GM-HM inequality.

AM–GM Inequality and Harmonic Mean: the key idea

For nonnegative a,b, $(\sqrt{a}-\sqrt{b})^{2}\ge 0$ gives $\frac{a+b}{2}\ge\sqrt{ab}$ , the two-term arithmetic–geometric mean inequality. The n-term version says the arithmetic mean is at least the geometric mean for nonnegative inputs, with equality when all inputs agree. For positive inputs the harmonic mean $H=\frac{n}{\sum 1/a_i}$ satisfies G≥H. The positivity matters for reciprocals. Choose terms so their product or sum matches the constraint, and check whether equal inputs are actually allowed. A proved bound is the minimum or maximum only if it is attained.

A worked example

Positive x,y have x+y=12. Find the greatest possible xy.

AM-GM gives $\sqrt{xy}\le\frac{x+y}{2}=6$ , so xy≤36. Equality requires x=y, and x=y=6 meets the constraint. Thus 36 is the maximum, not merely an upper bound.

Your turn: change one thing

For x>0, find the minimum of x+9/x.

Try this on paper before opening the explanation.

Compare your reasoning

The two positive terms have product 9, so their sum is at least $2\sqrt{9}=6$ . Equality requires x=9/x, giving x=3. This is allowed, so the minimum is 6.

Pause and check

A trap to avoid: Using negative inputs or giving a lower bound whose equality violates the constraint.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

For positive a,b prove their harmonic mean is at most their geometric mean.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

H=2ab/(a+b) and $G=\sqrt{ab}$ . Since $a+b\ge 2\sqrt{ab}>0$ , dividing 2ab by these positive denominators gives $H\le\frac{2ab}{2\sqrt{ab}}=\sqrt{ab}=G$ . Equality occurs exactly when a=b.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. AM-GM for two nonnegative numbers states which inequality?

Foundation

$a+b\le 2\sqrt{ab}$
a+b≥ab always
a+b=2ab
$a+b\ge 2\sqrt{ab}$

Hint

Recall the mean comparison.

Answer and reasoning

$a+b\ge 2\sqrt{ab}$ . The arithmetic mean (a+b)/2 is at least the geometric mean $\sqrt{ab}$ .

2. When is equality attained in two-term AM-GM?

Foundation

a+b=1 always
a or b is negative
a=b
a=2b

Hint

The proof uses $(\sqrt{a}-\sqrt{b})^{2}$ .

Answer and reasoning

a=b. The square vanishes exactly when $\sqrt{a}=\sqrt{b}$ , equivalently a=b.

3. Positive a+b=10. Find the maximum possible ab.

Core

Hint

Equalise the two terms.

Answer and reasoning

25. ab≤((a+b)/2)²=25, attained at a=b=5.

4. For x>0, find the minimum of x+16/x.

Core

Hint

The product of the two terms is 16.

Answer and reasoning

8. AM-GM gives at least $2\sqrt{16}=8$ , attained at x=4.

5. Why is checking equality necessary in an optimisation problem?

Stretch

Every inequality has strict equality
It replaces the proof
It only matters for integer arithmetic
A bound might not be attainable under the constraints

Hint

An infimum may differ from an attained minimum.

Answer and reasoning

A bound might not be attainable under the constraints. The equality conditions must correspond to allowed inputs before the bound can be claimed as a maximum or minimum.

6. For positive inputs, how are the three means ordered?

Stretch

Arithmetic ≥ geometric ≥ harmonic
Harmonic ≥ geometric ≥ arithmetic
Geometric ≥ arithmetic ≥ harmonic
All are always equal

Hint

Recall the two standard mean comparisons.

Answer and reasoning

Arithmetic ≥ geometric ≥ harmonic. AM≥GM≥HM, with equality throughout when all inputs agree.

7. For positive a,b, their geometric mean is what?

Foundation

$\sqrt{ab}$
(a+b)/2
2ab/(a+b)
ab

Hint

It is the square root of their product.

Answer and reasoning

$\sqrt{ab}$ . The two-input geometric mean is $\sqrt{ab}$ .

8. Positive x+y=14. Maximum xy?

Core

Hint

Use AM-GM with equal inputs.

Answer and reasoning

49. xy≤(14/2)²=49, attained at x=y=7.

9. Can AM-GM be used directly on x and −x for nonzero real x?

Stretch

No, both terms are not nonnegative
Yes always
Only for integer x
Only if x is large

Hint

Check the theorem’s domain.

Answer and reasoning

No, both terms are not nonnegative. One term is negative, so the usual nonnegative-input form does not apply.

Choose your next step

Continue to Weighted means. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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