AM-GM, Cauchy-Schwarz, and Power Mean
Pillar: FIX — "Fix a constraint, then the inequality tells you the extremum."
The Big Idea
Classical inequalities are tools for finding extreme values without calculus. The pattern is always the same:
- You have a constraint (e.g., \(xy = 36\)).
- You want to minimize or maximize some expression (e.g., \(x + y\)).
- An inequality gives you a bound on the expression.
- The equality condition tells you exactly when the bound is achieved.
This is the Fix pillar in action: fix the constraint, and the inequality forces the extremum. No derivatives, no Lagrange multipliers --- just algebra.
AM-GM for Two Variables
Theorem (AM-GM). For non-negative reals \(a, b\):
Equality holds if and only if \(a = b\).
In words: the arithmetic mean is at least the geometric mean.
Proof
Start from something obviously true:
Expand:
Rearrange:
Divide by 2. Done. Equality holds when \(\sqrt{a} = \sqrt{b}\), i.e., \(a = b\).
Why This Proof Matters
The proof reveals the engine behind AM-GM: it's a squared quantity \(\ge 0\). Every classical inequality ultimately reduces to this idea. When you see AM-GM applied, mentally picture a squared difference being non-negative.
Using AM-GM for Optimization
Example 1: Minimize a Sum Given a Product
Problem. Positive reals \(x, y\) satisfy \(xy = 36\). Find the minimum of \(x + y\).
Solution. By AM-GM:
So \(x + y \ge 12\). Equality when \(x = y = 6\).
The minimum value is \(\boxed{12}\).
Example 2: Maximize a Product Given a Sum
Problem. Positive reals \(x, y\) satisfy \(x + y = 10\). Find the maximum of \(xy\).
Solution. By AM-GM:
Equality when \(x = y = 5\). The maximum is \(\boxed{25}\).
The Dual Pattern
| Given | Find | AM-GM gives | Equality |
|---|---|---|---|
| Product \(= k\) | Minimize sum | \(\text{sum} \ge 2\sqrt{k}\) | When terms are equal |
| Sum \(= s\) | Maximize product | \(\text{product} \le (s/2)^2\) | When terms are equal |
This duality is worth memorizing. Fixed product, minimize sum. Fixed sum, maximize product. AM-GM handles both.
AM-GM for \(n\) Variables
Theorem. For non-negative reals \(a_1, a_2, \ldots, a_n\):
Equality if and only if \(a_1 = a_2 = \cdots = a_n\).
Example 3: Three Variables
Problem. Positive reals \(a, b, c\) satisfy \(abc = 27\). Minimize \(a + b + c\).
By AM-GM for three variables:
So \(a + b + c \ge 9\). Equality when \(a = b = c = 3\).
The Splitting Trick
Sometimes you need to apply AM-GM to terms that aren't naturally separated.
Problem. For \(x > 0\), minimize \(x + \frac{4}{x}\).
Apply AM-GM to the two terms \(x\) and \(\frac{4}{x}\):
Equality when \(x = \frac{4}{x}\), i.e., \(x = 2\).
Problem. For \(x > 0\), minimize \(x^2 + \frac{3}{x}\).
Direct AM-GM on two terms gives \(x^2 + \frac{3}{x} \ge 2\sqrt{3x}\), which still depends on \(x\). Not useful.
Instead, split \(\frac{3}{x}\) into two copies of \(\frac{3}{2x}\):
Equality when \(x^2 = \frac{3}{2x}\), i.e., \(x = \sqrt[3]{\frac{3}{2}}\).
Key insight: Split terms so that when you set all AM-GM arguments equal, the variable cancels inside the product. The number of copies controls the exponents in the equality condition.
Cauchy-Schwarz Inequality
Theorem. For real numbers \(a_1, \ldots, a_n\) and \(b_1, \ldots, b_n\):
Equality if and only if \(\frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_n}{b_n}\) (or some \(b_i = 0\) with corresponding \(a_i = 0\)).
Proof Sketch (Vectors)
Think of \(\mathbf{a} = (a_1, \ldots, a_n)\) and \(\mathbf{b} = (b_1, \ldots, b_n)\) as vectors. Then:
This is \(|\mathbf{a}||\mathbf{b}| \ge |\mathbf{a} \cdot \mathbf{b}|\), which says the dot product is at most the product of the magnitudes. Geometrically: \(\cos\theta \le 1\).
Example 4
Problem. If \(a^2 + b^2 = 1\), find the maximum of \(3a + 4b\).
By Cauchy-Schwarz:
Equality when \(\frac{a}{3} = \frac{b}{4}\), i.e., \(a = \frac{3}{5}, b = \frac{4}{5}\).
Maximum is \(\boxed{5}\).
Engel Form (Titu's Lemma)
The Engel form of Cauchy-Schwarz is the most useful version for competitions.
Theorem (Titu's Lemma). For positive reals \(a_i\) and positive reals \(b_i\):
Equality when \(\frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_n}{b_n}\).
Why This Form Is Powerful
It turns a sum of fractions into a single fraction. Whenever you see \(\sum \frac{(\text{something})^2}{\text{something else}}\), Titu's Lemma should be your first thought.
Example 5
Problem. Positive reals \(a, b, c\) satisfy \(a + b + c = 12\). Find the minimum of \(\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a}\).
By Titu's Lemma:
Equality when \(\frac{a}{b} = \frac{b}{c} = \frac{c}{a}\), which gives \(a = b = c = 4\).
Minimum is \(\boxed{12}\).
Example 6 (Competition Style)
Problem. For positive reals \(x, y, z\) with \(xyz = 1\), prove:
By Titu's Lemma:
By AM-GM for three variables: \(\frac{x + y + z}{3} \ge \sqrt[3]{xyz} = 1\), so \(x + y + z \ge 3\).
Therefore:
Power Mean Inequality
The Power Mean inequality unifies AM-GM and extends it to a whole family of means.
For positive reals \(a_1, \ldots, a_n\), define the power mean of order \(p\):
Special cases:
| \(p\) | Name | Formula (for \(n=2\)) |
|---|---|---|
| \(-1\) | Harmonic Mean (HM) | \(\frac{2}{\frac{1}{a} + \frac{1}{b}}\) |
| \(0\) (limit) | Geometric Mean (GM) | \(\sqrt{ab}\) |
| \(1\) | Arithmetic Mean (AM) | \(\frac{a+b}{2}\) |
| \(2\) | Quadratic Mean (QM) | \(\sqrt{\frac{a^2+b^2}{2}}\) |
Power Mean Inequality: If \(p < q\), then \(M_p \le M_q\). In particular:
Equality throughout if and only if \(a_1 = a_2 = \cdots = a_n\).
Numerical Check
Let \(a = 3, b = 12\).
- \(\text{HM} = \frac{2}{\frac{1}{3} + \frac{1}{12}} = \frac{2}{\frac{5}{12}} = \frac{24}{5} = 4.8\)
- \(\text{GM} = \sqrt{36} = 6\)
- \(\text{AM} = \frac{15}{2} = 7.5\)
- \(\text{QM} = \sqrt{\frac{9 + 144}{2}} = \sqrt{76.5} \approx 8.75\)
Indeed \(4.8 \le 6 \le 7.5 \le 8.75\).
When to Use Which Mean
| Situation | Mean | Why |
|---|---|---|
| Sum is fixed, bound product | AM \(\ge\) GM | Product \(\le (\text{AM})^n\) |
| Product is fixed, bound sum | GM \(\le\) AM | Sum \(\ge n \cdot \text{GM}\) |
| Sum of reciprocals appears | HM \(\le\) AM | Reciprocal sums are bounded by AM |
| Sum of squares appears | QM \(\ge\) AM | Squares bounded below by AM |
| Rates/speeds problems | HM | Average speed is harmonic mean of speeds |
Summary Table
| Inequality | Statement | Equality condition | Use when |
|---|---|---|---|
| AM-GM (\(n=2\)) | \(\frac{a+b}{2} \ge \sqrt{ab}\) | \(a = b\) | Sum-product optimization |
| AM-GM (\(n\) vars) | \(\text{AM} \ge \text{GM}\) | All equal | Multi-variable optimization |
| Cauchy-Schwarz | \((\sum a_i^2)(\sum b_i^2) \ge (\sum a_i b_i)^2\) | \(a_i/b_i\) constant | Max of linear expression given quadratic constraint |
| Titu's Lemma | \(\sum \frac{a_i^2}{b_i} \ge \frac{(\sum a_i)^2}{\sum b_i}\) | \(a_i/b_i\) constant | Sum of squared-over-linear terms |
| Power Mean | \(M_p \le M_q\) for \(p < q\) | All equal | Relating different types of averages |
Practice Problems
Problem 1. For \(a, b > 0\) with \(a + b = 1\), find the minimum of \(\frac{1}{a} + \frac{1}{b}\).
Problem 2. For \(x > 0\), find the minimum of \(x^3 + \frac{4}{x^3}\).
Problem 3. If \(a^2 + b^2 + c^2 = 1\), find the maximum of \(a + 2b + 3c\).
Problem 4. For positive reals \(a, b, c\), prove: \(\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \ge \frac{3}{2}\).
Problem 5. The harmonic mean of two positive numbers is \(4\) and their AM is \(9\). Find the two numbers.
Solution 1
By Titu's Lemma: \(\frac{1}{a} + \frac{1}{b} = \frac{1^2}{a} + \frac{1^2}{b} \ge \frac{(1+1)^2}{a+b} = \frac{4}{1} = 4\).
Equality when \(\frac{1}{a} = \frac{1}{b}\), i.e., \(a = b = \frac{1}{2}\). Minimum is \(4\).
Solution 2
By AM-GM: \(x^3 + \frac{4}{x^3} \ge 2\sqrt{x^3 \cdot \frac{4}{x^3}} = 2\sqrt{4} = 4\).
Equality when \(x^3 = \frac{4}{x^3}\), i.e., \(x^6 = 4\), \(x = \sqrt[6]{4} = 2^{1/3}\). Minimum is \(4\).
Solution 3
By Cauchy-Schwarz: \((a^2+b^2+c^2)(1^2+2^2+3^2) \ge (a+2b+3c)^2\).
\((1)(14) \ge (a+2b+3c)^2\), so \(a + 2b + 3c \le \sqrt{14}\).
Equality when \(\frac{a}{1} = \frac{b}{2} = \frac{c}{3} = t\). Then \(t^2 + 4t^2 + 9t^2 = 1\), \(14t^2 = 1\), \(t = \frac{1}{\sqrt{14}}\). Maximum is \(\sqrt{14}\).
Solution 4
Write \(\frac{a}{b+c} = \frac{a^2}{a(b+c)}\). By Titu's Lemma:
The denominator equals \(2(ab + bc + ca)\). By AM-GM, \((a+b+c)^2 \ge 3(ab+bc+ca)\).
So the RHS \(\ge \frac{3(ab+bc+ca)}{2(ab+bc+ca)} = \frac{3}{2}\).
Solution 5
Let the numbers be \(a, b\). We have \(\frac{2ab}{a+b} = 4\) and \(\frac{a+b}{2} = 9\).
From the second: \(a + b = 18\). From the first: \(ab = \frac{4 \cdot 18}{2} = 36\).
So \(a\) and \(b\) are roots of \(t^2 - 18t + 36 = 0\).
\(t = \frac{18 \pm \sqrt{324 - 144}}{2} = \frac{18 \pm \sqrt{180}}{2} = 9 \pm 3\sqrt{5}\).
The two numbers are \(9 + 3\sqrt{5}\) and \(9 - 3\sqrt{5}\).