Proofs of the Mean Inequalities

Try this

After each adjustment, has the “sum” changed? What about the “product”?
Each adjustment pulls one number exactly to the mean $A$ . How many adjustments do four numbers need in total?
Switch to “smoothing” (replace both numbers by their mean) and try a few times. Is the product still growing? Can all the numbers reach $A$ in finitely many steps?
When everything equals $A$ at the end, what is the product? Is it bigger or smaller than the product at the start?

Everything rests on a one-line identity. Let $x \leq A \leq y$ , and replace $x$ by $A$ and $y$ by $x + y - A$ (the sum is unchanged). The product changes by

A (x + y - A) - x y = (A - x) (y - A) \geq 0.

One factor is “how far the small number is from $A$ ”, the other is “how far the large number is from $A$ ”. When neither is zero, the product grows strictly.

Predict first

Take the four numbers $2, 4, 7, 11$ (mean $6$ ), and each time adjust the smallest and the largest. At most how many adjustments are needed before all four equal $6$ ?

2

3

4

It may never finish

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

Why: proving $A_{n} \geq G_{n}$ by adjustment

Adjustment: the sum is kept, the product never drops

Step 1 / 6

What we want to prove
Positive numbers $a_{1}, \dots, a_{n}$ have mean $A$ . We want $a_{1} a_{2} \dots a_{n} \leq A^{n}$ , that is, $G_{n} \leq A_{n}$ . Right now the product is $616$ , while $6^{4} = 1296$ .

$a_{1} a_{2} \dots a_{n} \leq A^{n}$
The smallest is at most A, the largest at least A
Adjust once
Adjust again
The last adjustment
When does equality hold?

Another route: forward–backward induction (Cauchy)

The book gives more than ten proofs. One of the most famous is Cauchy's “leapfrog” induction: first push $n = 2$ up to $4, 8, 16, \dots$ , then step back down one at a time.

2 → 4 → 8 → …, then back from n to n − 1

Step 1 / 5

Start: n = 2
$\frac{a + b}{2} \geq ab$ , proved at the previous stop.

$\frac{a + b}{2} \geq ab$
Doubling: 2 → 4
Keep doubling
Stepping back: n → n − 1
Tidy up

The trick of “adding one copy of the mean” will come up again and again: adding a term that does not change the mean is the standard way to pad out an inequality with the terms you need.

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: the product does not decrease after the adjustment

Each time you pick “one below $A$ and one above $A$ ”, the product can only grow.

Practice

Exercise 1Basic

Apply one adjustment to the three numbers $1, 5, 9$ (the smallest goes to the mean, the largest absorbs the difference). What is the product of the three numbers afterwards?

Exercise 2Basic

A list of positive numbers has mean $5$ , and two of the numbers are $3$ and $10$ . After one adjustment of these two, by how much has the product of these two numbers increased?

Exercise 3Advanced

Assume the mean inequality for four positive numbers. Using nothing else, deduce the mean inequality for three positive numbers $a, b, c$ : $\frac{a + b + c}{3} \geq 3 ab c$ .

Exercise 4Challenge

Positive numbers $x_{1}, \dots, x_{n}$ have product $1$ . Prove by induction that $x_{1} + \dots + x_{n} \geq n$ , and deduce $A_{n} \geq G_{n}$ from it.

Try this

Why: proving $A_{n} \geq G_{n}$ by adjustment

Adjustment: the sum is kept, the product never drops

Another route: forward–backward induction (Cauchy)

2 → 4 → 8 → …, then back from n to n − 1

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Averaging pairs and assuming everything becomes equal in finitely many steps

Proofs of the Mean Inequalities

Try this

Why: proving An​≥Gn​ by adjustment

Another route: forward–backward induction (Cauchy)

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Why: proving $A_{n} \geq G_{n}$ by adjustment