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

Algebra ->  Finance -> 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       Log On


   

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) About Me  (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


%28a%2Bb%29%2F2 >= sqrt%28ab%29  <------>  a + b >= 2%2Asqrt%28ab%29  <------>  %28sqrt%28a%29%29%5E2 + %28sqrt%28b%29%29%5E2  >= 2%2Asqrt%28a%29%2Asqrt%28b%29  <------>  %28sqrt%28a%29%29%5E2+-+2%2Asqrt%28a%29%2Asqrt%28b%29+%2B+%28sqrt%28b%29%5E2%29 >= 0  <------>  %28sqrt%28a%29+-+sqrt%28b%29%29%5E2 >= 0


It is true for all real non-negative  numbers  "a" and "b",   and the equality is hold if and only if   sqrt%28a%29 - sqrt%28b%29 = 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. 


sqrt%28ab%29 >= %282ab%29%2F%28a%2Bb%29  <------>  a + b%29 >= %282ab%29%2Fsqrt%28ab%29  <------>  a + b%29 >= 2%2Asqrt%28ab%29,


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

Solved.