Extrema with AM–GM · Little Blue Book, Visualized

Extrema with AM–GM

In the Chinese book: 用平均值不等式求极值

Positive, fixed, equal: split terms to fix the product, then check equality.

What this move does

The mean inequality links “sum” and “product”:

\frac{x _{1} + \dots + x _{n}}{n} \geq n x_{1} \dots x_{n} .

So when the product is fixed, the sum has a minimum; when the sum is fixed, the product has a maximum. To use it for extreme values, remember three checks: positive, fixed, equal.

Positive: every term involved is a positive number.
Fixed: by splitting terms and balancing coefficients, make the product (or the sum) of these terms a constant that does not depend on the variable.
Equal: the equality condition (all terms equal) can really be met within the range the problem allows. Otherwise all you have is a bound.

Predict first

Let $x > 0$ . What is the minimum of $x^{2} + \frac{16}{x}$ ?

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

Work through one

Example: the minimum of x² + 16/x (x > 0)

Step 1 / 5

Positive
$x > 0$ , so both terms are positive and AM–GM can be used.

$f (x) = x^{2} + \frac{16}{x}, x > 0$
Used directly: the product is not fixed
Fixed: split a term
Equal
The split is not unique

“A fixed sum gives the largest product” is the same fact seen the other way round. For a rectangle with fixed perimeter, the sum of the sides $a, b$ is fixed: the area is $ab = G^{2}$ , and $G \leq A$ .

Drag the split point

C

: the diameter (

a + b

) stays the same, and the half-chord

ab

is longest right in the middle, so the square has the largest area.

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: the bound from AM–GM is the minimum

$x + \frac{4}{x}$ ( $x > 0$ ): the product is the constant $4$ , so $x + \frac{4}{x} \geq 4$ , reached at $x = 2$ .

Practice

Exercise 1Basic

Let $x > 0$ . Find the minimum of $2 x + \frac{8}{x}$ .

Exercise 2Basic

Let $0 < x < \frac{1}{2}$ . Find the maximum of $x (1 - 2 x)$ .

Exercise 3Advanced

Let $x > 0$ . Find the minimum of $x^{2} + \frac{2}{x}$ .

Exercise 4Challenge

Let $x > 1$ . Find the minimum of $\frac{x ^{2} + 3}{x - 1}$ .

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

Using AM–GM twice with clashing equality conditions

a, b > 0

and

a + b = 1

. Since

a + \frac{1}{a} \geq 2

and

b + \frac{1}{b} \geq 2

, the minimum of

(a + \frac{1}{a}) (b + \frac{1}{b})

4

What went wrong? Think first, then open

Diagnosis The two inequalities need $a = 1$ and $b = 1$ for equality, but then $a + b = 2 \neq = 1$ . Equality cannot be reached, so $4$ is only a lower bound, not the minimum. In fact at $a = b = \frac{1}{2}$ the product is $(\frac{5}{2})^{2} = \frac{25}{4}$ , and that is the minimum.

Expand:

ab + \frac{1}{ab} + \frac{a}{b} + \frac{b}{a}

. Since

ab \leq \frac{1}{4}

and

t + \frac{1}{t}

is decreasing on

0 < t \leq \frac{1}{4}

, we get

ab + \frac{1}{ab} \geq \frac{17}{4}

; also

\frac{a}{b} + \frac{b}{a} \geq 2

. Both equalities hold at

a = b = \frac{1}{2}

, so they are compatible, and the minimum is

\frac{25}{4}

“Every “≥” must reach equality at the same point”