Question: Prove that the geometric mean is always less than or equal to the arithmetic mean (the arithmetic-geometric inequality) when r1 = 1 (the case described

Prove that the geometric mean is always less than or equal to the arithmetic mean (the arithmetic-geometric inequality) when r1 = 1 (the case described in Exercise 25).
The geometry behind the geometric mean is based on the following argument. If a random variable R takes on each of the values r1 and r2 with probability 0.5, a rectangle with sides of length r1 and r2 has area equal to that of a square with sides with length equal to the geometric mean.

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