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

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Question 982143: prove that the geometric mean of two positive, unequal numbers is less than their arithmetic mean.
Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!

Let  a  and  b  be two positive, unequal real numbers.

Then their arithmetic mean is  %28a+%2B+b%29%2F2  and geometric mean is  sqrt%28a%2Ab%29.

We need to prove that

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

Let us start with this inequality

%28sqrt%28a%29+-+sqrt%28b%29%29%5E2 > 0,

which is always true for any real unequal numbers  a  and  b,  because the left part is the square of the real number  sqrt%28a%29+-+sqrt%28b%29,  which is unequal to zero.

Rewrite this inequality in this way step by step:

a+-+2sqrt%28a%29%2Asqrt%28b%29+%2B+b > 0,         (apply the formula of the square of a difference)

a+%2B+b > 2sqrt%28a%29%2Asqrt%28b%29,             (move the term  -+2sqrt%28a%29%2Asqrt%28b%29  from the left side to the right with the opposite sign)

%28a%2Bb%29%2F2 > sqrt%28a%2Ab%29.

The last inequality is exactly what you have to prove.