SOLUTION: Prove that the geometric mean of two positive, unequal numbers is less than their arithmetic mean.

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Question 984031: Prove that the geometric mean of two positive, unequal numbers is less than their arithmetic mean.
Answer by ikleyn(52792) About Me  (Show Source):
You can put this solution on YOUR website!

Let  a  and  b  be two real positive unequal numbers.

Their arithmetic mean is  %28a+%2B+b%29%2F2.  Their geometric mean is  sqrt%28a%2Ab%29.  We need to prove that

sqrt%28a%2Ab%29 < %28a+%2B+b%29%2F2.

Let us start with inequality

%28sqrt%28a%29+-+sqrt%28b%29%29%5E2 > 0.             (1)

This inequality is true because the difference  sqrt%28a%29+-+sqrt%28b%29  is not zero and the square of a non-zero real number is positive.  Now,  expand  (1)  as

%28sqrt%28a%29+-+sqrt%28b%29%29%5E2 = %28sqrt%28a%29%29%5E2 - 2%2Asqrt%28a%29%2Asqrt%28b%29 + %28sqrt%28b%29%29%5E2 = a - 2%2Asqrt%28a%29%2Asqrt%28b%29 + b.

Thus the original inequality  (1)  is equivalent to

a - 2%2Asqrt%28a%29%2Asqrt%28b%29 + b > 0,     or

a + b > 2%2Asqrt%28a%29%2Asqrt%28b%29,     or

%28a+%2B+b%29%2F2 > sqrt%28a%2Ab%29.

It is exactly what has to be proved.