Important inequalities: the inequality for means

Ask the tutor about Important inequalities: the inequality for means Open the graphing workbench

Why do four different recipes all get to be called means? Sitting somewhere is not enough. Between the smaller and the larger of two positive numbers there is room for many values that nobody calls a mean. The four that survive have a fixed order.

These are means of positive numbers, not the partial sums on finite geometric and arithmetic series.

Two squares, not yet a mean

Nothing here is called a mean yet. Take two real numbers $a$ and $b$ . They may be negative or zero, and a sign does not change $∣ ab ∣$ .

Drag a or b. Both are real numbers, and a minus sign does not change |ab|. The upper mark is |ab|. The lower mark is (a²+b²)/2. The lower mark stays to the right until |a|=|b|, when the two marks share an x-coordinate.

Drag $a$ or $b$ . The upper mark is $∣ ab ∣$ , the lower mark is $\frac{1}{2} (a^{2} + b^{2})$ , and the lower mark stays to the right until $∣ a ∣ = ∣ b ∣$ . That meeting is equality. Setting only one of them to $0$ leaves the lower mark at half the square of the other, so the two marks do not share the origin.

For real $a$ and $b$ ,

$∣ ab ∣ \leq \frac{1}{2} (a^{2} + b^{2}) .$

Put $c = a^{2}$ and $d = b^{2}$ . Both are positive whenever $a$ and $b$ are nonzero. The geometric mean of $c$ and $d$ is $∣ ab ∣$ , and the arithmetic mean is $\frac{1}{2} (a^{2} + b^{2})$ . The inequality above is that one step of the next chain, written before the chain has a name.

Four marks between the ends

Now freeze two positive real numbers $c$ and $d$ . Positivity is the extra assumption: a reciprocal has to make sense, so neither number may be $0$ or negative. The numbers you drag stop above $0$ for that reason.

$\frac{2}{\frac{1}{c} + \frac{1}{d}} \leq c d \leq \frac{c + d}{2} \leq \frac{c ^{2} + d ^{2}}{2} .$

From left to right those are the harmonic mean, the geometric mean, the arithmetic mean, and the quadratic mean. Each of them lies between $min {c, d}$ and $max {c, d}$ . That is what earns the name mean.

Drag c or d. Both sliders stop above 0, because a nonpositive number leaves this chain. The segment runs from the smaller to the larger. On it, from left to right, sit the harmonic mean H, the geometric mean G, the arithmetic mean M, and the quadratic mean S. Set c equal to d and the four points meet.

Drag $c$ or $d$ . The segment runs from the smaller number to the larger. On it, left to right, sit $H$ , $G$ , $M$ , and $S$ . Set $c = d$ and the four points meet: the chain has collapsed to the only number available. Push one of them toward $0$ and $H$ drops with it, and $H$ still does not fall past the nearer end.

The same order for more than two

The two-number picture is the chain with the count frozen at $2$ . For positive real numbers $x_{1}, \dots, x_{n}$ ,

$min {x_{1}, \dots, x_{n}} \leq h \leq g \leq m \leq s \leq max {x_{1}, \dots, x_{n}},$

where

$m = \frac{x _{1} + \dots + x _{n}}{n} = \frac{1}{n} \sum_{k = 1}^{n} x_{k}$

is the arithmetic mean,

$g = (x_{1} x_{2} \dots x_{n})^{1/ n} = (\prod_{k = 1}^{n} x_{k})^{1/ n}$

is the geometric mean,

$h = \frac{n}{\frac{1}{x _{1}} + \dots + \frac{1}{x _{n}}}$

is the harmonic mean, and

$s = (\frac{1}{n} \sum_{k = 1}^{n} x_{k}^{2})^{1/2}$

is the quadratic mean. The chain runs $h$ , then $g$ , then $m$ , then $s$ , so the names are defined in that order.

Drag a, b, or c. Each slider stops above 0, and the count is frozen at three. The segment runs from the smallest to the largest. H, G, M, and S stay in that order on it. Set the three sliders equal and the four points meet.

The board freezes the count at three. Drag $a$ , $b$ , or $c$ . The same four points stay in the same order between the smallest and the largest. Set the three numbers equal and they meet, as they did for two.

The boards check that order, for two positive numbers and for three. This page does not prove the chain.

Drop positivity and the harmonic mean, which divides by the numbers, leaves the hypothesis of this page. The length comparison for complex numbers is a different inequality, on the triangle inequality. The line under $(1 + x)^{n}$ does not produce this order; that comparison is Bernoulli's inequality.