SOLUTION: We may define the three means (arithmetic, geometric, and harmonic) of two positive numbers a and b as A = (a+b)/2 G = √ab H = 2ab/ a+b, respectively a. Show the

Question 1120745: We may define the three means (arithmetic, geometric, and harmonic) of two positive numbers a and b as
A = (a+b)/2
G = √ab
H = 2ab/ a+b, respectively
a. Show the inequality, A ≥ G ≥ H, holds if and only if a = b
Answer by ikleyn(52800) (Show Source): You can put this solution on YOUR website!
.

1. Prove A >= G holds if and only if a = b.

You have this chain of equivalent inequalities


 >=   <------>  a + b >=   <------>   +   >=   <------>   >= 0  <------>   >= 0


It is true for all real non-negative  numbers  "a" and "b",   and the equality is hold if and only if    -  = 0, 


which in turn is equivalent to  a = b.


You can read this chain of inequalities / (of arguments) from the right to the left - then you will get the proof you need.

2. Prove G >= H holds if and only if a = b.

 >=   <------>   +  >=   <------>   +  >= ,


and the last inequality is exactly  THE SAME  as the proven in  #1.

Solved.

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SOLUTION: We may define the three means (arithmetic, geometric, and harmonic) of two positive numbers a and b as A = (a+b)/2 G = &#8730;ab H = 2ab/ a+b, respectively a. Show the

SOLUTION: We may define the three means (arithmetic, geometric, and harmonic) of two positive numbers a and b as A = (a+b)/2 G = √ab H = 2ab/ a+b, respectively a. Show the