# 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, =√(ab)=8…(2)
Equations (1) and (2) can be written as

a+b=20…(1)
ab=64…(2)

Putting the value of a and b from (1), (2) in the identity (ab)2=(a+b)2-4ab, we get
(ab)2=400-256=144
or ab=±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. =√ab=8→ab=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. √ab=5⇒ab=25…(2)
The quadratic equation is given by,

x2x (Sum of roots)+(Product of roots)=0
x2x(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., Multiplying the three inequalities, we get 📌 Example 5. The minimum Value of the expression 3x+3(1-x), xR, is

(A) 0 (B) ⅓ (C) 3 (D) 2√3
✍ Solution: (D) is the correct answer. We know A.M.≥G.M. for positive numbers.
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