Inequalities path · I06

Before this lesson: Arithmetic, geometric and harmonic means, Arithmetic and exact calculation

Your goal: Use nonnegative weights with their sum explicitly normalised.

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: weighted arithmetic–geometric mean inequality.

Weighted AM–GM Inequality: the key idea

Weights describe how much each input contributes. For positive x,y and 0≤t≤1, weighted AM-GM states tx+(1−t)y≥xty1−ttx+(1-t)y\ge x^{t}y^{1-t}. The weights must be nonnegative and add to 1. Rational weights follow by repeating terms in ordinary AM-GM: weights 1/3 and 2/3 mean one copy of x and two copies of y. Real weights follow by continuity from rational weights. If both weights are positive, equality requires x=y; a zero weight imposes no restriction on the omitted input. Matching the weights to the exponents often removes a variable from the product.

A worked example

For x>0, prove x+2x≥3x+\frac2{\sqrt{x}}\ge3.

Apply three-term AM-GM to x,1x\frac1{\sqrt{x}},1x\frac1{\sqrt{x}}. Their product is 1, so their sum is at least 3. Equality means x=1xx=\frac1{\sqrt{x}}, hence x=1.

Your turn: change one thing

For a,b>0 prove a+3b≥4(ab3)1/4a+3b\ge 4(a b^{3})^{1/4}.

Try this on paper before opening the explanation.

Compare your reasoning

Apply four-term AM-GM to a,b,b,b. Equality holds exactly when a=b.

Pause and check

A trap to avoid: Weights not summing to one or a zero input with an undefined power.

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 weighted AM-GM with weights 2/5 and 3/5 for positive a,b.

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

Apply five-term AM-GM to a,a,b,b,b. Their arithmetic mean is (2a+3b)/5 and their geometric mean is (a2b3)1/5=a2/5b3/5(a^{2}b^{3})^{1/5}=a^{2/5}b^{3/5}. Equality holds 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. Which weights are valid for weighted AM-GM?

Foundation

  1. 1/4 and 3/4
  2. −1 and 2
  3. 1 and 1
  4. 2 and 3
Hint

Check nonnegativity and total.

Answer and reasoning

1/4 and 3/4. Both weights are nonnegative and sum to 1.

2. Weights 1/3 and 2/3 correspond to which repeated list?

Foundation

  1. a,a,b,b
  2. a,b,b
  3. a,a,b
  4. a,b
Hint

Count the relative frequencies.

Answer and reasoning

a,b,b. One a and two b give the required weights.

3. For x>0, the minimum of x+2xx+\frac2{\sqrt{x}} is what?

Core

  1. 3
  2. 2
  3. 4
  4. 0
Hint

Use three positive terms.

Answer and reasoning

3. AM-GM gives 3, attained at x=1.

4. For positive a,b, when is a+3b=4(ab3)1/4a+3b=4(a b^{3})^{1/4}?

Core

  1. a=b
  2. a=3b
  3. a=0
  4. Always
Hint

Equality requires the repeated terms to match.

Answer and reasoning

a=b. The four terms a,b,b,b must be equal.

5. How can rational weights m/n and (n−m)/n be handled?

Stretch

  1. Assume m=n
  2. Repeat the inputs m and n−m times
  3. Subtract the inputs
  4. Ignore the denominator
Hint

Use n-term AM-GM.

Answer and reasoning

Repeat the inputs m and n−m times. The repeated list has exactly the required arithmetic and geometric means.

6. If a weight is zero, what does equality require of its input?

Stretch

  1. It must be one
  2. It must equal all other inputs
  3. No restriction from that term
  4. It must be zero
Hint

The input is omitted from both means.

Answer and reasoning

No restriction from that term. Zero-weight inputs do not affect the comparison.

7. In weighted AM-GM, the weights add to what?

Foundation

  1. 2
  2. The number of inputs
  3. 1
  4. 0
Hint

Normalise the weights.

Answer and reasoning

1. The weighted mean uses nonnegative weights totalling 1.

8. For positive x, the minimum of x+3x1/3x+\frac3{x^{1/3}} is what?

Core

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

Use four terms x and three copies of x−1/3x^{-1/3}.

Answer and reasoning

4. Their product is 1, so their sum is at least 4, attained at x=1.

9. Why repeat b three times when proving a+3b≥4(ab3)1/4a+3b\ge 4(a b^{3})^{1/4}?

Stretch

  1. To make b vanish
  2. Because a must equal 3b
  3. Because there are only two terms
  4. Its multiplicity creates the exponent 3/4
Hint

The geometric mean multiplies all repeated terms.

Answer and reasoning

Its multiplicity creates the exponent 3/4. Four-term AM-GM uses a,b,b,b, whose product is ab³.

Choose your next step

Continue to Power mean inequality. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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