How to Prove the Quadratic Mean–Arithmetic Mean Inequality

The quadratic mean–arithmetic mean inequality says that the root mean square of a list of numbers is at least its arithmetic mean. The standard proof is a direct application of the Cauchy–Schwarz inequality. It works for any finite list of real numbers, and gives the slightly stronger statement that the quadratic mean is at least the absolute value of the arithmetic mean.

State the inequality

For real numbers $x_1,x_2,\ldots,x_n$, define the quadratic mean (QM) and arithmetic mean (AM) by

$$\operatorname{QM}=\sqrt{\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}}, \qquad \operatorname{AM}=\frac{x_1+x_2+\cdots+x_n}{n}.$$

Then

$$\operatorname{QM}\geq |\operatorname{AM}|\geq \operatorname{AM}.$$

When the inputs are nonnegative, this is usually written simply as $\operatorname{QM}\geq\operatorname{AM}$. Nonnegativity is often assumed in statements about means, but it is not needed for the inequality itself.

Proof using Cauchy–Schwarz

Apply Cauchy–Schwarz to the vectors $(x_1,x_2,\ldots,x_n)$ and $(1,1,\ldots,1)$. It gives

$$(x_1+x_2+\cdots+x_n)^2 \leq (x_1^2+x_2^2+\cdots+x_n^2)(1^2+1^2+\cdots+1^2).$$

The second factor on the right is $n$, so

$$(x_1+x_2+\cdots+x_n)^2 \leq n(x_1^2+x_2^2+\cdots+x_n^2).$$

Divide both sides by $n^2$:

$$\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^2 \leq \frac{x_1^2+x_2^2+\cdots+x_n^2}{n}.$$

The left side is $\operatorname{AM}^2$ and the right side is $\operatorname{QM}^2$. Since $\operatorname{QM}$ is nonnegative, taking square roots yields

$$\operatorname{QM}\geq |\operatorname{AM}|.$$

Finally, $|\operatorname{AM}|\geq\operatorname{AM}$ for every real number $\operatorname{AM}$. Therefore $\operatorname{QM}\geq\operatorname{AM}$.

A proof by sums of squares

For nonnegative $x_i$, there is also a useful proof that makes the gap between the two means visible. The identity

$$n\sum_{i=1}^n x_i^2-\left(\sum_{i=1}^n x_i\right)^2 =\sum_{1\leq i<j\leq n}(x_i-x_j)^2$$

has a nonnegative right side. Hence $n\sum x_i^2\geq(\sum x_i)^2$. Dividing by $n^2$ and taking square roots gives $\operatorname{QM}\geq\operatorname{AM}$.

To verify the identity, expand each square on the right. Each $x_i^2$ appears in exactly $n-1$ pairs, while each product $x_ix_j$ appears with coefficient $-2$. The result is $(n-1)\sum x_i^2-2\sum_{i<j}x_ix_j$, which is also the expansion of the left side.

Worked example

Take the numbers $1,2,5$. Their arithmetic mean is

$$\operatorname{AM}=\frac{1+2+5}{3}=\frac{8}{3}\approx 2.67.$$

Their quadratic mean is

$$\operatorname{QM}=\sqrt{\frac{1^2+2^2+5^2}{3}} =\sqrt{10}\approx 3.16.$$

Thus $\operatorname{QM}>\operatorname{AM}$. The sum-of-squares identity also shows why equality fails here:

$$(1-2)^2+(1-5)^2+(2-5)^2=1+16+9=26>0.$$

At the level of squared means, the difference is $10-(8/3)^2=26/9$, consistent with the identity after dividing by $n^2=9$.

When does equality hold?

Equality in Cauchy–Schwarz occurs when the two vectors are proportional. Here that means $(x_1,\ldots,x_n)$ is a constant multiple of $(1,\ldots,1)$, so all the inputs are equal.

For nonnegative inputs, all-equal inputs give equality: if each $x_i=c\geq0$, then both means equal $c$. For arbitrary real inputs, the stronger inequality $\operatorname{QM}\geq|\operatorname{AM}|$ is an equality whenever all inputs are equal. Equality in the weaker statement $\operatorname{QM}\geq\operatorname{AM}$ holds exactly when all inputs are equal to the same nonnegative number. If the arithmetic mean is negative, equality is impossible because $\operatorname{QM}\geq0>\operatorname{AM}$; if the arithmetic mean is zero, equality forces every input to be zero.

Common mistakes

• Forgetting the absolute value for signed inputs. The direct Cauchy–Schwarz result is $\operatorname{QM}\geq|\operatorname{AM}|$. The usual AM statement follows because $|\operatorname{AM}|\geq\operatorname{AM}$.
• Taking square roots before dividing correctly. The inequality is between $\operatorname{AM}^2$ and $\operatorname{QM}^2$ only after dividing the sum-of-squares inequality by $n^2$.
• Claiming equality for any list with the same average. Equality in Cauchy–Schwarz requires the entries themselves to be equal, not merely their average to have a particular value.
• Mixing up QM with the mean of absolute values. The quadratic mean uses $x_i^2$ inside the average and then takes one square root; it is not $\frac{1}{n}\sum |x_i|$.

The same argument applies to weighted means when the weights are nonnegative and sum to one: the weighted quadratic mean is at least the weighted arithmetic mean. This follows by applying Cauchy–Schwarz to the vectors with components $\sqrt{w_i}x_i$ and $\sqrt{w_i}$.