QM-AM-GM-HM inequalities

In mathematics, the QM-AM-GM-HM inequalities, also known as the mean inequality chain, state the relationship between the harmonic mean, geometric mean, arithmetic mean, and quadratic mean (also known as root mean square). Suppose that ${\displaystyle x_{1},x_{2},\ldots ,x_{n}}$ are positive real numbers. Then

${\displaystyle 0<{\frac {n}{{\frac {1}{x_{1}}}+{\frac {1}{x_{2}}}+\cdots +{\frac {1}{x_{n}}}}}\leq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}\leq {\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\leq {\sqrt {\frac {x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}}{n}}}.}$^[1]

In other words, QM≥AM≥GM≥HM. These inequalities often appear in mathematical competitions and have applications in many fields of science.^{[citation needed]}

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, ...):

${\displaystyle \left(\sum _{i=1}^{n}x_{i}\cdot 1\right)^{\!2}\leq \left(\sum _{i=1}^{n}x_{i}^{2}\right)\left(\sum _{i=1}^{n}1^{2}\right)=n\,\sum _{i=1}^{n}x_{i}^{2},}$ hence ${\displaystyle \left({\frac {\sum _{i=1}^{n}x_{i}}{n}}\right)^{\!2}\leq {\frac {\sum _{i=1}^{n}x_{i}^{2}}{n}}}$. For positive ${\displaystyle x_{i}}$ the square root of this gives the inequality.

HM-GM inequality

The reciprocal of the harmonic mean is the arithmetic mean of the reciprocals ${\displaystyle 1/x_{1},\dots ,1/x_{n}}$, and it exceeds ${\displaystyle 1/{\sqrt[{n}]{x_{1}\dots x_{n}}}}$ by the AM-GM inequality. ${\displaystyle x_{i}>0}$ implies the inequality:

${\displaystyle {\frac {n}{{\frac {1}{x_{1}}}+\dots +{\frac {1}{x_{n}}}}}\leq {\sqrt[{n}]{x_{1}\dots x_{n}}}.}$^[2]

The n = 2 case

The semi-circle used to visualize the inequalities

When n = 2, the inequalities become

${\displaystyle {\frac {2x_{1}x_{2}}{x_{1}+x_{2}}}\leq {\sqrt {x_{1}x_{2}}}\leq {\frac {x_{1}+x_{2}}{2}}\leq {\sqrt {\frac {x_{1}^{2}+x_{2}^{2}}{2}}}}$ for all ${\displaystyle x_{1},x_{2}>0,}$ ^[3]

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.

Comparison of harmonic, geometric, arithmetic, quadratic and other mean values of two positive real numbers ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$

See also

• Inequalities among pythagorean means
• Generalized mean inequality