[10/2023] HM-GM-AM-QM Inequalities

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Mathematical relationships

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 x1,x2,…,xn{displaystyle x_{1},x_{2},ldots ,x_{n}} are positive real numbers. Then

0<n1x1+1×2+⋯+1xn≤x1x2⋯xnn≤x1+x2+⋯+xnn≤x12+x22+⋯+xn2n.{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}}}.}

These inequalities often appear in mathematical competitions and have applications in many fields of science.

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, …):

(∑i=1nxi⋅1)2≤(∑i=1nxi2)(∑i=1n12)=n∑i=1nxi2,{displaystyle left(sum _{i=1}^{n}x_{i}cdot 1right)^{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 (∑i=1nxin)2≤∑i=1nxi2n{displaystyle left({frac {sum _{i=1}^{n}x_{i}}{n}}right)^{2}leq {frac {sum _{i=1}^{n}x_{i}^{2}}{n}}}. For positive xi{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 1/x1,…,1/xn{displaystyle 1/x_{1},dots ,1/x_{n}}, and it exceeds 1/x1…xnn{displaystyle 1/{sqrt[{n}]{x_{1}dots x_{n}}}} by the AM-GM inequality. xi>0{displaystyle x_{i}>0} implies the inequality:

n1x1+⋯+1xn≤x1…xnn.{displaystyle {frac {n}{{frac {1}{x_{1}}}+dots +{frac {1}{x_{n}}}}}leq {sqrt[{n}]{x_{1}dots x_{n}}}.}

The n = 2 case


The semi-circle used to visualize the inequalities

When n = 2, the inequalities become

21×1+1×2≤x1x2≤x1+x22≤x12+x222{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}}}} for all x1,x2>0,{displaystyle x_{1},x_{2}>0,}

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

x1{displaystyle x_{1}}


x2{displaystyle x_{2}}

To infer the correct order, the four expressions can be evaluated with two small numbers.

For x1=10{displaystyle x_{1}=10} and x2=40{displaystyle x_{2}=40} in particular, this results in 16<20<25<301−118{displaystyle 16<20<25<30{sqrt {1-{tfrac {1}{18}}}}}.

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