Proving Inequalities with AM–GM

What this move does

The two sides of an inequality to be proved are often polynomials of the same degree, for example

a^{3} + b^{3} + c^{3} \geq a^{2} b + b^{2} c + c^{2} a (a, b, c \geq 0) .

The mean inequality says: the geometric mean of a few numbers (built from their product) is at most their arithmetic mean. Think of each term as its vector of exponents; the exponents of the geometric mean are then the average of the exponents. So:

For each term on the small side (a target), find a few terms on the big side (sources) such that the target's exponents are a weighted average of the sources' exponents. In the exponent triangle: the target lies in the region spanned by the sources.
Find the weights $w_{i}$ and write the weighted mean inequality $target \leq \sum w_{i} \cdot source_{i}$ . With a common denominator, this is an ordinary $n$ -term AM–GM (with some sources repeated).
Cycle $a \to b \to c \to a$ and add the three inequalities. Finally check the equality condition: all sources equal.

Predict first

To write $a^{2} b \leq x a^{3} + y b^{3}$ (with $x + y = 1$ ), what should $x$ be?

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

Work through one

Example: prove a³ + b³ + c³ ≥ a²b + b²c + c²a, and its fourth-degree sibling

Step 1 / 7

Where is the target?
The exponents of $a^{2} b$ are $(2, 1, 0)$ : on the bottom edge, near the $a^{3}$ corner. The big side has only the three corners $a^{3}, b^{3}, c^{3}$ .

$a^{2} b \leq ?$
a³ alone is not enough
Add b³
Add the cyclic versions
Equality
Degree four: a target inside the triangle
Add cyclically, done

The exponent triangle is more than a helper picture: it tells you directly which inequalities can be proved this way. If the target lies outside the sources' region, weighted AM–GM is powerless (and such an “inequality” is usually false in the first place).

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: AM–GM bounds the target by the sources

$ab c$ sits right in the middle of the small triangle formed by $a^{2} b$ , $b^{2} c$ , $c^{2} a$ : $ab c \leq \frac{1}{3} (a^{2} b + b^{2} c + c^{2} a)$ , which is just the three-term mean inequality.

Practice

Exercise 1Basic

Write $a b^{2} \leq x a^{3} + y b^{3}$ (with $x + y = 1$ and $a, b \geq 0$ ). What is $x$ ?

Exercise 2Basic

Prove that $a^{2} + b^{2} + c^{2} \geq ab + b c + c a$ for all real numbers $a, b, c$ , and say when equality holds.

Exercise 3Advanced

Let $a, b, c \geq 0$ . Prove that $a^{4} + b^{4} + c^{4} \geq a^{3} b + b^{3} c + c^{3} a$ .

Exercise 4Challenge

Let $x, y, z > 0$ . Prove that $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq \frac{x + y + z}{3 x y z}$ .

What this move does

Work through one

Example: prove a³ + b³ + c³ ≥ a²b + b²c + c²a, and its fourth-degree sibling

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Taking “half and half” as the weights without thinking