Inequalities path · I09

Before this lesson: Rearrangement inequality

Your goal: Recognise when ordering permits a product-of-averages comparison.

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: Chebyshev sum inequality; Chebyshev inequality for ordered sequences.

Chebyshev’s Sum Inequality: the key idea

Chebyshev’s sum inequality compares the average of products with the product of averages. For similarly ordered real lists, ∑aibin≥∑ain∑bin\frac{\sum a_i b_i}{n}\ge\frac{\sum a_i}{n}\frac{\sum b_i}{n}. Oppositely ordered lists reverse the inequality. Do not apply it to unsorted unrelated pairings. A useful identity is n∑aibi−(∑ai)(∑bi)=∑i<j(aj−ai)(bj−bi)n\sum a_{i}b_{i}-(\sum a_{i})(\sum b_{i})=\sum_{i<j}(a_{j}-a_{i})(b_{j}-b_{i}). Every term on the right is nonnegative for similarly ordered lists. If both lists are strictly increasing and n>1, the inequality is strict.

A worked example

Use Chebyshev on the lists 1,2,3 and 1,4,9.

Both increase. The average of products is (1+8+27)/3=12. The product of averages is 2·14/3=28/3. Thus 12≥28/3.

Your turn: change one thing

If a≤b and c≤d, compare 2(ac+bd) and (a+b)(c+d).

Try this on paper before opening the explanation.

Compare your reasoning

Their difference is (b−a)(d−c)≥0, the two-term Chebyshev inequality.

Pause and check

A trap to avoid: Applying the same direction to oppositely ordered sequences.

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

Prove Chebyshev for two similarly ordered real lists of equal length.

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

Expand ∑i<j(aj−ai)(bj−bi)\sum_{i<j}(a_{j}-a_{i})(b_{j}-b_{i}). Each aᵢbᵢ appears n−1 times and each off-diagonal product appears once with a minus sign. This equals n∑aibi−(∑ai)(∑bi)n\sum a_{i}b_{i}-(\sum a_{i})(\sum b_{i}). Each product of differences is nonnegative. Divide by n² to obtain the mean form.

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. Chebyshev in its usual direction needs which condition?

Foundation

  1. Equal lengths are unnecessary
  2. Similarly ordered lists
  3. Positive integers only
  4. Equal sums
Hint

The products of differences must be nonnegative.

Answer and reasoning

Similarly ordered lists. Both lists must have equal length and the same ordering direction.

2. For oppositely ordered lists, what happens to the inequality?

Foundation

  1. It becomes undefined
  2. It becomes equality always
  3. It reverses
  4. It stays strict in the same direction
Hint

The difference products are nonpositive.

Answer and reasoning

It reverses. The mean product is then at most the product of means.

3. For lists 1,3 and 2,4, what is the mean of paired products?

Core

  1. 8
  2. 14
  3. 7
  4. 6
Hint

Add products and divide by 2.

Answer and reasoning

7. (1·2+3·4)/2=7.

4. For the same lists, what is the product of means?

Core

  1. 3
  2. 6
  3. 7
  4. 14
Hint

The means are 2 and 3.

Answer and reasoning

6. 2·3=6, which is at most 7.

5. Which identity supports the n-term proof?

Stretch

  1. ∑aibi=0\sum a_{i}b_{i}=0
  2. n∑aibi<0n\sum a_{i}b_{i}<0 always
  3. n∑aibi−∑ai∑bi=∑i<j(aj−ai)(bj−bi)n\sum a_{i}b_{i}-\sum a_{i}\sum b_{i}=\sum_{i<j}(a_{j}-a_{i})(b_{j}-b_{i})
  4. ∑aibi=∑ai∑bi\sum a_{i}b_{i}=\sum a_{i}\sum b_{i} always
Hint

Expand the pair differences.

Answer and reasoning

n∑aibi−∑ai∑bi=∑i<j(aj−ai)(bj−bi)n\sum a_{i}b_{i}-\sum a_{i}\sum b_{i}=\sum_{i<j}(a_{j}-a_{i})(b_{j}-b_{i}). Diagonal terms and off-diagonal terms give exactly the displayed identity.

6. If both lists strictly increase and n>1, is equality possible?

Stretch

  1. Always
  2. Only for positive inputs
  3. Only for n=2
  4. No
Hint

At least one pair difference product is positive.

Answer and reasoning

No. Every pair difference product is positive, so the total is positive.

7. Chebyshev compares which quantities?

Foundation

  1. Average of paired products and product of averages
  2. Only sums of angles
  3. Only maximum and minimum entries
  4. Only factorials
Hint

Read its mean form.

Answer and reasoning

Average of paired products and product of averages. It compares ∑aibin\frac{\sum a_i b_i}{n} with ∑ain∑bin\frac{\sum a_i}{n}\frac{\sum b_i}{n}.

8. For 1,2 and 4,8, the average of products is what?

Core

  1. 9
  2. 12
  3. 20
  4. 10
Hint

Compute (1·4+2·8)/2.

Answer and reasoning

10. (4+16)/2=10.

9. If one list is constant, what happens in Chebyshev?

Stretch

  1. It is always strict
  2. The inequality fails
  3. The other list must vanish
  4. Equality holds
Hint

Factor the constant out of each side.

Answer and reasoning

Equality holds. Both sides equal that constant times the average of the other list.

Choose your next step

Continue to Cauchy–Schwarz inequality. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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