Arithmetic Mean ≥ Geometric Mean ≥ Harmonic Mean (AM ≥ GM ≥ HM)

Relationship among different means: AM≥GM≥HM
where, A.M=Arithmatic mean, G.M=Geomatric mean and H.M=Harmonic mean
💎 Proof:
Let us consider two observation x₁ and x₂. Then

i,e. AM>GM.
The equality sign holds good, if x₁=x₂.

Again

i,e. GM>HM
The equality sign holds good, if x₁=x₂.
∴ AM≥GM≥HM
The theorem is true for any number of observations.

Relationship Between A.M. and GM.
Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively. Then

From (i), we obtain the relationship A≥G.

📌 Example 1. If A.M. and GM. of two positive numbers a and b are 10 and 8, respectively, find the numbers.
✍ Solution:
Given that A.M. =½(a+b)=10…(1)
and G, M, =√(a⋅b)=8…(2)
Equations (1) and (2) can be written as

a+b=20…(1)
a⋅b=64…(2)
Putting the value of a and b from (1), (2) in the identity (a–b)²=(a+b)²-4ab, we get
(a–b)²=400-256=144
or a–b=±12 … (3)
Solving (1) and (3), we obtain
a=4, b=16 or a=16, b=4
Thus, the numbers a and b are 4, 16 or 16, 4 respectively.

📌 Ex2. The A.M between two numbers is 10 and their G.M is 8. Determine the numbers.
✍ Solution:

A.M =½(a+b)=10→a+b=20…(1)

G.M. =√a⋅b=8→a⋅b=64…(2)

from (2), b=64/a, put in (1)

Hence the numbers are 4 and 16.

📌 Ex3. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
✍ Solution:
Let the root of the quadratic equation be a and b.
According to the given condition,
A.M. ½(a+b)=8⇒a+b=16…(1)
G.M. √a⋅b=5⇒a⋅b=25…(2)
The quadratic equation is given by,

x²–x (Sum of roots)+(Product of roots)=0
x²–x(a+b)+(ab)=0
x²-16x+25=0 [Using (1) and (2)]
Thus, the required quadratic equation is x²-16x+25=0.

Objective Type Questions
Choose the correct answer out of the four given options in Examples 4 and 5.
📌 Example 4. If x, y, z are positive integers then value of expression (x+y)(y+z)(z+x) is

(A) =8xyz (B) >8xyz (C) <8xyz (D) =4xyz
✍ Solution: (B) is the correct answer, since A.M. > G.M.,