Power Mean

M_r ≤ M_s for r ≤ s — one inequality unifies HM, GM, AM, and QM.

The idea

Definition (Power mean).

For positive reals $a_1, \dots, a_n$ and any real $r \neq 0$, the $r$-th power mean is $M_r = \left(\frac{1}{n}\sum_{i=1}^{n} a_i^{\,r}\right)^{1/r},$ and $M_0$ is defined as the limiting value as $r \to 0$, which is the geometric mean $(a_1 \cdots a_n)^{1/n}$.

Theorem (Power mean inequality).

For positive reals $a_1, \dots, a_n$ and real numbers $r \le s$, $M_r \le M_s,$ and for $r < s$ equality holds if and only if $a_1 = \cdots = a_n$.

The familiar means belong to this family: the harmonic, geometric, arithmetic and quadratic means are $M_{-1}$, $M_0$, $M_1$ and $M_2$. The single statement $M_r \le M_s$ therefore contains every comparison between them, including AM-GM as the case $M_0 \le M_1$.

Larger $r$ gives a larger mean because raising the entries to a high power exaggerates their differences: the largest entries dominate the sum, and taking the $1/r$-th root afterwards does not undo that bias. As $r \to \infty$, $M_r$ approaches the maximum entry, and as $r \to -\infty$ it approaches the minimum. A list of equal entries has no differences to exaggerate, and exactly then all the power means agree.

Ways to work on it

• Walkthrough. Identify the four classical means in the power-mean family, then verify AM ≤ QM.
• Proof. See why the power means increase with their exponent — one application of Jensen's inequality.
• Practice. Compute AM and QM for a random pair and check AM ≤ QM.
• Hardest. Compute HM and GM for a random triple and verify HM ≤ GM.

Not sure where to start? Take the ten-question placement test.