Variants of AM–GM · Little Blue Book, Visualized

What this move does

Apply the mean inequality to numbers that have been “transformed”, and you get a whole family of new inequalities:

min a_{i} \leq H \leq G \leq A \leq Q \leq max a_{i} .

They are all special cases ( $p = - 1, 0, 1, 2$ ) of the power mean $M_{p} = (\frac{a _{1}^{p} + \dots + a _{n}^{p}}{n})^{1/ p}$ , and $M_{p}$ increases as $p$ increases. The most useful variants:

Reciprocals: applying $A \geq G$ to the $\frac{1}{a _{i}}$ gives $H \leq G$ . Together, $A \geq H$ , that is, $(\sum a_{i}) (\sum \frac{1}{a _{i}}) \geq n^{2}$ .
Squares: $Q \geq A$ , that is, $n \sum a_{i}^{2} \geq (\sum a_{i})^{2}$ . When the sum is fixed, the sum of squares has a minimum.
Weights: $w_{1} a_{1} + \dots + w_{n} a_{n} \geq a_{1}^{w_{1}} \dots a_{n}^{w_{n}}$ ( $w_{i} > 0$ , $\sum w_{i} = 1$ ). The “multiply by $1$ ” trick is using exactly this.

Predict first

Positive numbers $a, b, c$ satisfy $a + b + c = 3$ . What is the minimum of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ ?

Choose an answer first; then you can see the result.

Work through one

A walk along the power mean curve

Step 1 / 6

p = −1: the harmonic mean
$M_{- 1} = \frac{3}{1 + \frac{1}{2} + \frac{1}{6}} = \frac{3}{5/3} = 1.8$ . It is the reciprocal of “the arithmetic mean of the reciprocals”.

$H = \frac{n}{\sum \frac{1}{a _{i}}}$
Substituting reciprocals: H ≤ G
p = 1: the arithmetic mean
p = 2: the quadratic mean
The two ends: minimum and maximum
Equality

Practice

Exercise 1Basic

Positive numbers $a, b, c$ satisfy $a + b + c = 6$ . Find the minimum of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ .

Exercise 2Basic

Real numbers $x, y$ satisfy $x + y = 4$ . Find the minimum of $x^{2} + y^{2}$ .

Exercise 3Advanced

Positive numbers $a, b$ satisfy $a + b = 1$ . Find the minimum of $\frac{1}{a} + \frac{9}{b}$ .

Exercise 4Challenge

Let $a, b, c > 0$ . Prove that $(a + b + c) (\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a}) \geq \frac{9}{2}$ .

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Minimizing 1/a + 4/b under a + b = 1 with two AM–GMs in a row

\frac{1}{a} + \frac{4}{b} \geq 2 \frac{4}{ab} = \frac{4}{ab} \geq \frac{4}{\frac{a + b}{2}} = 8

, so the minimum is

8

What went wrong? Think first, then open

Diagnosis The first equality needs $\frac{1}{a} = \frac{4}{b}$ , that is, $b = 4 a$ ; the second needs $a = b$ . Both hold together only at $a = b = 0$ , which is impossible. $8$ is never reached; it is only a lower bound.

Multiply by

1 = a + b

(a + b) (\frac{1}{a} + \frac{4}{b}) = 5 + \frac{b}{a} + \frac{4 a}{b} \geq 5 + 4 = 9

, with equality at

b = 2 a

, that is,

a = \frac{1}{3}

b = \frac{2}{3}

, which is allowed. The minimum is

9

“Get there with one application; do not let two equality cases fight”