The QM–AM–GM–HM inequalities for three positive numbers

2026-10-22 huynhvietkhanh Comments: 0

Prove the three-number chain of means and apply it with normalization, products, reciprocals, and complete equality conditions.

High schoolInequalities

Content updated: October 22, 2026

For three positive numbers, the chain of four means remains valid. Its proof requires a separate three-number AM–GM argument, followed by applying it to reciprocals to compare the geometric and harmonic means.

Learning goals: prove the three-number chain, apply two- or three-number inequalities appropriately, use normalization assumptions, and determine equality.

Prerequisites: the four means for two numbers and the cubic algebraic identity.

On this page: Definition · Method · Examples · Practice · Further learning

1. Four definitions

Let $a,b,c>0$. Define

$$QM=\sqrt{\frac{a^2+b^2+c^2}{3}},\quad AM=\frac{a+b+c}{3},\quad GM=\sqrt[3]{abc},\quad HM=\frac{3}{1/a+1/b+1/c}.$$

All four are positive and satisfy $QM\ge AM\ge GM\ge HM$. Equality at each step, and throughout the chain, holds exactly when $a=b=c$. The harmonic mean is not the sum of reciprocals; the numerator $3$ is the number of entries.

2. Proving the three-number chain

Since the means are nonnegative, $QM\ge AM$ follows from

$$QM^2-AM^2=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{9}\ge0.$$

To prove $AM\ge GM$, set $u=\sqrt[3]a$, $v=\sqrt[3]b$, and $w=\sqrt[3]c$. Then

$$a+b+c-3\sqrt[3]{abc}=\frac{u+v+w}{2}\bigl((u-v)^2+(v-w)^2+(w-u)^2\bigr)\ge0.$$

This is the identity for $u^3+v^3+w^3-3uvw$. The sum $u+v+w$ is positive, so equality holds exactly when $u=v=w$, or $a=b=c$.

Apply $AM\ge GM$ to $1/a,1/b,1/c$ to obtain

$$\frac{1/a+1/b+1/c}3\ge\frac1{\sqrt[3]{abc}}.$$

Taking reciprocals of the positive sides reverses the inequality, giving $HM\le GM$. Equality requires equal reciprocals, which is again $a=b=c$.

In problems with $x+y+z=1$, use that assumption at the final step to replace the sum by $1$. When combining estimates, factors being multiplied must be nonnegative, and every equality case must also satisfy the prescribed sum.

3. Three worked examples

Example 1. A product of three sums

For $x,y,z>0$, prove $(x+1/y)(y+1/z)(z+1/x)\ge8$. Apply the two-number inequality to each sum:

$$x+\frac1y\ge2\sqrt{\frac xy},\quad y+\frac1z\ge2\sqrt{\frac yz},\quad z+\frac1x\ge2\sqrt{\frac zx}.$$

Multiplying the three positive inequalities gives right side $8$, since the product under the radical is $1$. Equality requires $xy=yz=zx=1$. Dividing the first two equations by $y>0$ gives $x=z$; combining this with $xz=1$ and positivity yields $x=z=1$ and then $y=1$.

Example 2. Three fractions with sum normalized to 1

Let $x,y,z>0$ and $x+y+z=1$. Set $U=xy/z$, $V=yz/x$, and $W=zx/y$. The pairs satisfy $U+V\ge2y$, $V+W\ge2z$, and $W+U\ge2x$, since $UV=y^2$, $VW=z^2$, and $WU=x^2$. Add and divide by $2$:

$$\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}=\frac{(U+V)+(V+W)+(W+U)}2\ge x+y+z=1.$$

Equality requires $U=V=W$, equivalent to $x=y=z$ because the variables are positive. Combining this with sum $1$ gives $x=y=z=1/3$.

Example 3. The sum of pairwise products

Again let $x,y,z>0$ and $x+y+z=1$. Apply the three-number inequality to $xy,yz,zx$ and to $x,y,z$:

$$(xy+yz+zx)(x+y+z)\ge3\sqrt[3]{(xy)(yz)(zx)}\cdot3\sqrt[3]{xyz}=9xyz.$$

Since the sum is $1$, we obtain $xy+yz+zx\ge9xyz$. Both steps are equalities at $x=y=z=1/3$. Conversely, equality in the application to $x,y,z$ forces the three numbers to be equal, so there are no other equality cases.

4. Practice

Let $a,b,c>0$ and $a+b+c=6$. Find the maximum value of $abc$.
For $a,b,c>0$, prove $1/a+1/b+1/c\ge9/(a+b+c)$.
Why does Example 1 have equality only at $(1,1,1)$ rather than at every triple with $x=y=z>0$?

Hints and answers

$abc\le((a+b+c)/3)^3=8$, attained at $a=b=c=2$.
Starting from $AM\ge HM$, multiply by the positive sums to obtain the result. Equality occurs when $a=b=c$.
If all three numbers equal $t>0$, the equality conditions still require $t=1/t$, so $t=1$. Equality of the three variables alone is insufficient.

5. Common mistakes

Do not replace the three-number proof by “similarly” without establishing the needed result. Do not apply a sum condition $1$ to a different triple of variables. After multiplying inequalities, solve all equality conditions; Example 1 also imposes a product equal to $1$.

6. Further learning

Basic properties of inequalities (English lesson) · Basic inequalities (English lesson) · The QM–AM–GM–HM inequalities for two numbers (English lesson).