AM–GM in Geometric Inequalities

In the Chinese book: 几何不等式中的平均值

Tangent lengths x, y, z sum to p; AM–GM on them, equality when equilateral.

What this move does

The three sides $a, b, c$ of a triangle cannot be chosen freely (they must satisfy the triangle inequality), so using AM–GM on them directly is often awkward. The incircle offers a beautiful substitution: the tangent lengths from the three vertices,

x = p - a, y = p - b, z = p - c, a = y + z, b = z + x, c = x + y .

$x, y, z$ can be any positive numbers, and $x + y + z = p$ (the semi-perimeter). So:

Replace the sides by tangent lengths: $a = y + z$ , $b = z + x$ , $c = x + y$ . The triangle inequality then holds automatically.
Write the quantities to compare in terms of $x, y, z$ : Heron's formula $S^{2} = p x y z$ , $r = \frac{S}{p}$ , $R = \frac{ab c}{4 S}$ .
Use AM–GM on $x, y, z$ ; the equality case $x = y = z$ is exactly the equilateral triangle.

Predict first

What is the largest possible area of a triangle with perimeter $6$ ? (Give three decimal places.)

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

Work through one

Two geometric inequalities, one mean inequality

Step 1 / 6

Tangent lengths
The two tangents from a vertex to the incircle have equal length. Each color appears twice, and together they make up the perimeter, so $x + y + z = p$ .

$2 (x + y + z) = a + b + c = 2 p$
Heron's formula in new clothes
Fixed sum, largest product
Equality: the equilateral triangle
Another inequality: R ≥ 2r
Translate into radii

Both inequalities reach equality at the equilateral triangle. That is no coincidence: after the tangent-length substitution, a symmetric inequality almost always has its equality case at $x = y = z$ .

Practice

Exercise 1Basic

For a triangle with perimeter $12$ , what is the largest possible value of the squared area $S^{2}$ ?

Exercise 2Basic

A right triangle has hypotenuse $10$ . What is its largest possible area?

Exercise 3Advanced

Prove that in every triangle $ab c \geq 8 (p - a) (p - b) (p - c)$ , and say when equality holds.

Exercise 4Challenge

Prove that in every triangle $a^{2} + b^{2} + c^{2} \geq 43 S$ .

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

A chain of inequalities, each with its own equality case

For a triangle with perimeter

6

S = \frac{1}{2} ab sin C \leq \frac{1}{2} ab \leq \frac{1}{2} (\frac{a + b}{2})^{2} < \frac{1}{2} \cdot 3^{2} = \frac{9}{2}

, so the maximum area is

\frac{9}{2}

What went wrong? Think first, then open

Diagnosis Every step is true, but the three equalities need $C = 9 0^{\circ}$ , $a = b$ and $a + b = 6$ (that is, $c = 0$ ) respectively. They cannot all hold at once, so $\frac{9}{2}$ is never reached. The true maximum is $3 \approx 1.73$ , less than half of that.

Switch to the tangent lengths

x, y, z

x + y + z = p = 3

and

S = 3 x y z \leq 3 \cdot 1^{3} = 3

, with equality at

x = y = z = 1

, the equilateral triangle with side

2

. This one is reached.

“In a chain of bounds, all the equality cases must hold at once, in the same figure”