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?


1 Answer 1


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 $x_1 > 0, x_2 = \ldots x_n = 0$.

  • $\begingroup$ Oh yes. Of course! Thanks. $\endgroup$
    – Auslander
    Commented Mar 26, 2017 at 11:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .