Inequalities path · I02

Before this lesson: Inequality rules and signs, Mathematical induction: the first principle

Your goal: State the domain and prove a product bound from a two-factor case.

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.

Weierstrass product inequalities: the key idea

Product bounds often begin by expanding two factors. If x,y≥0, then (1+x)(1+y)=1+x+y+xy≥1+x+y. Repeating the argument proves ∏(1+xi)≥1+∑xi\prod (1+x_{i})\ge 1+\sum x_{i} for nonnegative inputs. A related form is ∏(1−xi)≥1−∑xi\prod (1-x_{i})\ge 1-\sum x_{i} when 0≤xᵢ≤1. For the second form, multiplication by 1−xᵢ preserves order because it is nonnegative. These bounds are useful when a long product looks harder than a short sum. Check the sign hypotheses before using either form.

A worked example

Give a lower bound for (1+1/10)(1+1/20)(1+1/30).

All three increments are nonnegative, so the product is at least 1+1/10+1/20+1/30=71/60. The extra pairwise and triple products are nonnegative, which explains why the linear expression is a lower bound.

Your turn: change one thing

Prove (1−a)(1−b)≥1−a−b for 0≤a,b≤1.

Try this on paper before opening the explanation.

Compare your reasoning

Expand: (1−a)(1−b)−(1−a−b)=ab≥0. Equality holds exactly when ab=0.

Pause and check

A trap to avoid: Using a product inequality outside its hypotheses.

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 ∏i=1n(1+xi)≥1+∑i=1nxi\prod _{i=1}^{n}(1+x_{i})\ge 1+\sum _{i=1}^{n}x_{i} for xᵢ≥0 by induction.

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

For n=1 equality holds. Suppose the product of the first n factors is at least 1+S, where S is their nonnegative sum. Multiply by 1+xₙ₊₁>0: the new product is at least (1+S)(1+xₙ₊₁)=1+S+xₙ₊₁+Sxₙ₊₁≥1+S+xₙ₊₁. This proves the next case. Equality requires at most one positive input.

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. For x,y≥0, what is (1+x)(1+y)−(1+x+y)?

Foundation

  1. −xy
  2. 1
  3. xy
  4. x+y
Hint

Expand both sides.

Answer and reasoning

xy. The constant and linear terms cancel, leaving xy.

2. Which hypothesis guarantees xy≥0 here?

Foundation

  1. x≠y alone
  2. x,y≥0
  3. x<0<y
  4. x+y>0 alone
Hint

The sign of a product depends on both factors.

Answer and reasoning

x,y≥0. Two nonnegative factors have a nonnegative product.

3. For a,b≥0, which bound is valid?

Core

  1. (1+a)(1+b)≤1+a+b
  2. (1+a)(1+b)=1+a+b always
  3. (1+a)(1+b)<1
  4. (1+a)(1+b)≥1+a+b
Hint

Inspect the extra term ab.

Answer and reasoning

(1+a)(1+b)≥1+a+b. Expansion adds the nonnegative term ab.

4. For 0≤a,b≤1, when is (1−a)(1−b)=1−a−b?

Core

  1. When ab=0
  2. Only when a=b=1
  3. Whenever a=b
  4. Never
Hint

The difference is ab.

Answer and reasoning

When ab=0. Equality means the nonnegative difference ab vanishes.

5. Why require 1−xᵢ≥0 in the inductive proof of the minus-form bound?

Stretch

  1. To avoid all equality cases
  2. To ensure n=1
  3. To preserve an inequality when multiplying
  4. To make every sum prime
Hint

Track the multiplying factor.

Answer and reasoning

To preserve an inequality when multiplying. A negative multiplier would reverse the inequality and invalidate that step.

6. With all xᵢ≥0, when can ∏(1+xi)=1+∑xi\prod (1+x_{i})=1+\sum x_{i} hold?

Stretch

  1. All are strictly positive and n>1
  2. Only when n is even
  3. At most one xᵢ is positive
  4. Exactly two are positive
Hint

Look at the pairwise product terms.

Answer and reasoning

At most one xᵢ is positive. If two inputs are positive, their pairwise product makes the expansion strictly larger. If at most one is positive, every higher-order product vanishes.

7. For nonnegative x,y, which product expansion is correct?

Foundation

  1. (1+x)(1+y)=1+x+y+xy
  2. 1+x+y only
  3. 1+xy only
  4. 1−x−y+xy
Hint

Multiply the two brackets.

Answer and reasoning

(1+x)(1+y)=1+x+y+xy. Each first-factor term multiplies each second-factor term.

8. For x=1,y=2, by how much does (1+x)(1+y) exceed 1+x+y?

Core

  1. 3
  2. 6
  3. 2
  4. 1
Hint

The difference is xy.

Answer and reasoning

2. The product is 6 and the linear expression is 4.

9. Why is nonnegativity relevant to the product bound?

Stretch

  1. It makes the discarded higher products nonnegative
  2. It makes all inputs equal
  3. It makes multiplication unnecessary
  4. It restricts inputs to integers
Hint

Inspect xy in the two-factor case.

Answer and reasoning

It makes the discarded higher products nonnegative. Removing a nonnegative product can only decrease the expression.

Choose your next step

Continue to Absolute-value inequalities. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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