Hölder’s inequality: Examples and Practice

Before this lesson: Cauchy–Schwarz inequality, Weighted means

Your goal: Select matching factors and exponents instead of recalling a formula blindly.

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: Holder inequality; Hoelder inequality.

Hölder’s inequality: the key idea

For nonnegative lists aᵢ,bᵢ,cᵢ, Hölder gives $(\sum a_{i}^{3})(\sum b_{i}^{3})(\sum c_{i}^{3})\ge (\sum a_{i}b_{i}c_{i})^{3}$ . More generally, for p,q>1 with 1/p+1/q=1, $\sum|x_i y_i|\le\left(\sum|x_i|^p\right)^{1/p}\left(\sum|y_i|^q\right)^{1/q}$ . Cauchy is the case p=q=2. Match the exponents before applying the theorem. A useful consequence for positive yᵢ,zᵢ and nonnegative xᵢ is $\sum\frac{x_i^3}{y_i z_i}\ge\frac{(\sum x_i)^3}{(\sum y_i)(\sum z_i)}$ . This lesson applies Hölder and derives that consequence; the general theorem is stated as an available tool.

A worked example

For positive a,b,c prove a³+b³+c³≥(a+b+c)³/9.

Use three-list Hölder with lists (a,b,c),(1,1,1),(1,1,1). It gives (a³+b³+c³)·3·3≥(a+b+c)³. Equality requires a=b=c.

Your turn: change one thing

Positive a+b+c=3. Find the minimum of a³+b³+c³.

Try this on paper before opening the explanation.

Compare your reasoning

The previous bound gives at least 27/9=3, attained at a=b=c=1.

Pause and check

A trap to avoid: Mismatched exponents or unverified positivity.

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

Derive the displayed fraction form of Hölder.

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

Use lists $\frac{x_i}{(y_i z_i)^{1/3}}$ , $y_{i}^{1/3}$ and $z_{i}^{1/3}$ . Their termwise product is xᵢ. Their cube sums are respectively $\sum\frac{x_i^3}{y_i z_i}$ , $\sum y_{i}$ and $\sum z_{i}$ . Hölder gives the result after division by the two positive sums.

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. In two-list Hölder, p and q satisfy what?

Foundation

p+q=1
pq=1
p=q always
1/p+1/q=1

Hint

They are conjugate exponents.

Answer and reasoning

1/p+1/q=1. The reciprocal exponents add to 1, with each exponent greater than 1.

2. If p=3, what is q?

Foundation

Hint

Solve 1/3+1/q=1.

Answer and reasoning

3/2. 1/q=2/3, so q=3/2.

3. Positive a+b+c=3. Minimum of a³+b³+c³?

Core

Hint

Use three-list Hölder.

Answer and reasoning

3. The sum is at least 3³/9=3, attained at equal inputs.

4. The coefficient in (a+b+c)³≤k(a³+b³+c³) can be what sharp value?

Core

Hint

Test equal inputs after applying Hölder.

Answer and reasoning

9. Hölder gives 9; a=b=c=1 makes 27=9·3.

5. What does p=q=2 reduce Hölder to?

Stretch

Pigeonhole
Cauchy–Schwarz
Triangle congruence
Wilson’s theorem

Hint

Compare the squared sums.

Answer and reasoning

Cauchy–Schwarz. The two-list formula becomes the usual Cauchy bound.

6. In the fraction form, why must yᵢ,zᵢ be positive?

Stretch

Equality would otherwise always hold
They occur as denominators and positive cube-root factors
All variables must be integers
Negative cubes do not exist

Hint

Inspect the substitution.

Answer and reasoning

They occur as denominators and positive cube-root factors. The stated nonnegative-list application and division require positive denominators.

7. Three-list Hölder with cube sums uses which input condition in this lesson?

Foundation

Negative entries only
Prime entries
Distinct entries
Nonnegative entries

Hint

The stated product-cube form uses nonnegative lists.

Answer and reasoning

Nonnegative entries. Its hypotheses must match the version being applied.

8. If p=4 in two-list Hölder, what is the conjugate q?

Core

Hint

Solve 1/4+1/q=1.

Answer and reasoning

4/3. 1/q=3/4, so q=4/3.

9. Why verify equality after using Hölder in optimisation?

Stretch

Every bound is attained
Only integer inputs matter
The bound must be attainable under the constraint
Hölder has no equality cases

Hint

A sharp minimum needs an allowed equality configuration.

Answer and reasoning

The bound must be attainable under the constraint. A lower bound alone does not prove the minimum equals that bound.

Choose your next step

Continue to Inequalities in geometry. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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