Algebra theory AM GM HM Inequality

For n positive numbers a1,a2...an where n >= 1, their Arithmetic mean is >= their geometric mean.
i.e.
(a1+a2...an)/n >= (a1.a2.a3...an)^(1/n)

The proof is simple for 2 numbers. Then it's easy to extend for 2,8,16...2^n numbers.
Induction can be used for a general proof.

Proof for 2 positive numbers:
(a - b)^2 > =0
a^2 -2ab + b^2 >= 0
a^2 + 2ab + b^2 >= 4ab
(a+b) >= 2.sqrt(ab)
(a+b)/2 >= sqrt(ab)
AM >= GM

For 4 numbers(and similarly for any 2^n numbers):
Keep extending the 2 number result:
1/2{(a+b)/2 + (c+d)/2} >= (abcd)^(1/4)
and so on..


Ex1:
What is the minimum value of x + 1/x where x > 0.
Solution:
By AM GM inequality:
x + 1/x >= 2.(x.1/x)^(1/2)
x + 1/x >= 2
So 2 is the minimum value.

Ex2:
What is the maximum value of x + 1/x where x < 0?
Solution:
Let x = -p where p > 0
p + 1/p >= 2 by AM GM inequality.
-(p + 1/p) <= -2
(x + 1/x) <= -2
Answer: -2

Ex3:
Find the valid interval  for 1+ x + 1/x where x is a non-zero real number.
Solution:
If x > 0 then
(1 + x + 1/x) >= 3.(1.x.1/x)^1/3
So 1 + x + 1/x >= 3.1
1 + x + 1/x >= 3

If x < 0 then:
let p = -x for p > 0
p + 1/p >= 2
x + 1/x <= -2
1 + x + 1/x <= -1

So answer is (-Inf, -1] U [3, Inf).
___________________________________________

It can be further extended to Harmonic mean also:

The AM-GM-HM inequality refers to relationships between three classical means of positive real numbers: the Arithmetic Mean (AM), the Geometric Mean (GM), and the Harmonic Mean (HM). The inequality states:

AM≥GM≥HM\text{AM} \geq \text{GM} \geq \text{HM}

for any set of positive real numbers, with equality if and only if all the numbers are equal.

Definitions (for nn positive real numbers a1,a2,…,ana_1, a_2, \dots, a_n):

  • Arithmetic Mean (AM):

    AM=a1+a2+⋯+ann\text{AM} = \frac{a_1 + a_2 + \dots + a_n}{n}

  • Geometric Mean (GM):

    GM=(a1a2…an)1/n\text{GM} = \left(a_1 a_2 \dots a_n\right)^{1/n}

  • Harmonic Mean (HM):

    HM=n1a1+1a2+⋯+1an\text{HM} = \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}}

Inequality Statement:

a1+a2+⋯+ann≥(a1a2…an)1/n≥n1a1+1a2+⋯+1an\frac{a_1 + a_2 + \dots + a_n}{n} \geq \left(a_1 a_2 \dots a_n\right)^{1/n} \geq \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}}

Equality holds if and only if a1=a2=⋯=ana_1 = a_2 = \dots = a_n.

Example (with two numbers):

Let a=4a = 4, b=9b = 9:

  • AM=4+92=6.5\text{AM} = \frac{4 + 9}{2} = 6.5

  • GM=4⋅9=36=6\text{GM} = \sqrt{4 \cdot 9} = \sqrt{36} = 6

  • HM=214+19=21336=7213≈5.54\text{HM} = \frac{2}{\frac{1}{4} + \frac{1}{9}} = \frac{2}{\frac{13}{36}} = \frac{72}{13} \approx 5.54

So AM≈6.5≥6=GM≥HM≈5.54\text{AM} \approx 6.5 \geq 6 = \text{GM} \geq \text{HM} \approx 5.54


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