HM-GM-AM-QM inequalities
In mathematics, the HM-GM-AM-QM inequalities state the relationship between the harmonic mean, geometric mean, arithmetic mean, and quadratic mean (aka root mean square or RMS for short). Suppose that [math]\displaystyle{ x_1, x_2, \ldots, x_n }[/math] are positive real numbers. Then
- [math]\displaystyle{ 0\lt \frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}}\leq\sqrt[n]{x_1x_2\cdots x_n}\leq\frac{x_1+x_2+\cdots+x_n}{n} \leq\sqrt{\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}}. }[/math]
These inequalities often appear in mathematical competitions and have applications in many fields of science.
Proof
There are three inequalities between means to prove. There are various methods to prove the inequalities, including mathematical induction, the Cauchy–Schwarz inequality, Lagrange multipliers, and Jensen's inequality. For several proofs that GM ≤ AM, see Inequality of arithmetic and geometric means.
AM-QM inequality
From the Cauchy–Schwarz inequality on real numbers, setting one vector to (1, 1, ...):
- [math]\displaystyle{ \left( \sum_{i=1}^n x_i \cdot 1 \right)^2 \leq \left( \sum_{i=1}^n x_i^2 \right) \left( \sum_{i=1}^n 1^2 \right) = n \,\sum_{i=1}^n x_i^2, }[/math] hence [math]\displaystyle{ \left( \frac{\sum_{i=1}^n x_i}{n} \right)^2 \leq \frac{\sum_{i=1}^n x_i^2}{n} }[/math]. For positive [math]\displaystyle{ x_i }[/math] the square root of this gives the inequality.
HM-GM inequality
The reciprocal of the harmonic mean is the arithmetic mean of the reciprocals [math]\displaystyle{ 1/x_1, \dots, 1/x_n }[/math], and it exceeds [math]\displaystyle{ 1/\sqrt[n]{x_1 \dots x_n} }[/math] by the AM-GM inequality. [math]\displaystyle{ x_i \gt 0 }[/math] implies the inequality:
- [math]\displaystyle{ \frac{n}{\frac{1}{x_1} + \dots + \frac{1}{x_n}} \leq \sqrt[n]{x_1\dots x_n}. }[/math]
The n = 2 case
When n = 2, the inequalities become
- [math]\displaystyle{ \frac 2 {\frac{1}{x_1}+\frac{1}{x_2}} \leq \sqrt{x_1 x_2} \leq \frac{x_1+x_2}{2}\leq\sqrt{\frac{x_1^2+x_2^2}{2}} }[/math] for all [math]\displaystyle{ x_1, x_2 \gt 0, }[/math]
which can be visualized in a semi-circle whose diameter is [AB] and center D.
Suppose AC = x1 and BC = x2. Construct perpendiculars to [AB] at D and C respectively. Join [CE] and [DF] and further construct a perpendicular [CG] to [DF] at G. Then the length of GF can be calculated to be the harmonic mean, CF to be the geometric mean, DE to be the arithmetic mean, and CE to be the quadratic mean. The inequalities then follow easily by the Pythagorean theorem.
Tests
To infer the correct order, the four expressions can be evaluated with two small numbers.
For [math]\displaystyle{ x_1=10 }[/math] and [math]\displaystyle{ x_2=40 }[/math] in particular, this results in [math]\displaystyle{ 16 \lt 20 \lt 25 \lt 30 \sqrt{1 - \tfrac{1}{18}} }[/math].
External links
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