Inequalities:
A.M. ≥ G. M. ≥ H. M.:
Suppose a1, a2,... . . . , an be n positive real numbers, then it may be defined that A ≥ G ≥ H. Moreover equality holds at either place if and only if a1 = a2 = ....... = an.
Weighted Means:
Consider a1, a2, a3 . . . , an be n positive real numbers and m1, m2, . .. . . . , mn will be n positive rational numbers. Then we may described weighted Arithmetic mean (A*), weighted Geometric mean (G*) and weighted harmonic mean (H*) as

It may be defined that A* ≥ G* ≥ H*. Moreover equality holds at either place if and only if a1 = a2 = . . . . . = an.
Arithmetic Mean of mth Power:
Suppose a1, a2,......, an be n positive real numbers (not all equal) and let m be a natural number, then
if m ∈ R -[0, 1].
Yet if m ∈ (0, 1) , then 
Clearly if m∈{0, 1} , then 
Problem: Prove that x +1/x ≥ 2, if x > 0 and x + 1/x ≤ -2 , if x < 0 .
Solution: Since
( A.M. ≥ G. M. )
=> x+1/x ≥ 2.
If x < 0 , let y = -x , then y > 0
and 
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