The RMS-AM inequality states that for positive real numbers $x_1,\ldots,x_n$, $$AM=\frac{x_1+\cdots+x_n}{n}\leq\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}}=RMS.$$ For two positive numbers $x_1,x_2$, the inequality can be inferred geometrically from the diagram below.

enter image description here

Unless I'm mistaken, the picture also seems to imply that for two positive numbers $x_1,x_2$,

$$RMS^2\leq 2\cdot AM^2$$

so that $$AM=\frac{x_1+x_2}{2}\geq\frac{1}{\sqrt2}\sqrt{\frac{x_1^2+x_2^2}{2}}=\frac{1}{\sqrt2}RMS.$$

Is anyone aware of how this might generalise for more than two numbers?


For non-negative numbers $x_1,\ldots, x_n$ $$ x_1^2 + \ldots + x_n^2 \le (x_1 + \ldots + x_n)^2 $$ holds, as can be seen by expanding the right-hand side. It follows that $$ RMS^2 \le n \cdot AM^2 $$ or $$ AM \ge \frac{1}{\sqrt n} RMS $$ which generalizes your result for $n=2$.

The factor $\frac{1}{\sqrt n}$ is best possible because equality holds if $x_1 > 0, x_2 = \ldots x_n = 0$.

| cite | improve this answer | |
  • $\begingroup$ Oh yes. Of course! Thanks. $\endgroup$ – Auslander Mar 26 '17 at 11:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.