QM-AM-GM-HM inequalities explained

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

x1,x2,\ldots,xn

are positive real numbers. Then
0<n
1
+1
x2
+ … +1
xn
x1

\leq\sqrt[n]{x1x2 …

x\leq\sqrt{
n}\leqx1+x2+ … +xn
n
2
x
n
n
}.[1]

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 :

\left(

n
\sum
i=1

xi1\right)2\leq\left(

n
\sum
i=1
2
x
i

\right)\left(

n
\sum
i=1

12\right)=n

n
\sum
i=1
2,
x
i
hence

\left(

n
\sumxi
i=1
n

\right)2\leq

n
\sum
2
x
i
i=1
n
. For positive

xi

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

, and it exceeds

1/\sqrt[n]{x1...xn}

by the AM-GM inequality.

xi>0

implies the inequality:
n
1+...+
1
xn
x1

\leq\sqrt[n]{x1...xn}.

The n&thinsp;=&thinsp;2 case

When n&thinsp;=&thinsp;2, the inequalities become

2
1
+1
x2
x1

\leq\sqrt{x1x2}\leq

x1+x2\leq\sqrt{
2
2
x
2
2
} for all

x1,x2>0,

which can be visualized in a semi-circle whose diameter is [''AB''] and center D.

Suppose AC&thinsp;=&thinsp;x1 and BC&thinsp;=&thinsp;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 positive numbers.

For

x1=10

and

x2=40

in particular, this results in

16<20<25<5\sqrt{34}

.

See also

External links

Notes and References

  1. Book: Djukić, Dušan . The IMO compendium: a collection of problems suggested for the International Mathematical Olympiads, 1959-2009 . 2011 . Springer . International mathematical olympiad . 978-1-4419-9854-5 . Problem books in mathematics . New York . 7.