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Taking part in the 4th Malaysian Memory Championship for the third time, Chia wanted to test his mental arithmetic skill against three-time competition ...

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Prove that the geometric average is always smaller than the arithmetic average?
Q. Suppose you are given a set of n numbers. The arithmetic average is what you get when you sum all the numbers and divide by n. The geometric average is what you get when you multiply them all, then take the nth root. So how do you prove that for a set of positive numbers, the geometric average is always smaller than the geometric average? I mean for ANY whole n. I already know how to prove it works for n = 2 and n = 3, but how do you prove it works for any whole n.
Asked by chess_is_best - Sat Apr 18 10:23:15 2009 - - 3 Answers - 0 Comments

A. I'm going to prove this using Jensen's inequality. We have E(lnX)<=lnE(X) by Jensen's inequality, Where E(X) is the expected value of X. And E(lnX) is the expected value of lnX Therefore, E(lnX) =(lnX1+lnX2+...+ln(Xn))/n <=lnE(X) =ln((X1+X2+...Xn)/n) therefore (lnX1+lnX2+...+ln(Xn))/n <=ln((X1+X2+...Xn)/n) Now we apply the exponential function to both sides to get: e^((lnX1+lnX2+...+ln(Xn)) /n) =(X1*X2*...*Xn)^(1/n) <=e^(ln((X1+X2+...Xn)/n)) =(X1+...+Xn)/n Therefore, (X1*X2*...*Xn)^(1/n)<=(X1 +...+Xn)/n Which proves the geometric mean is always less than the arithmetic average.
Answered by Sam B - Sat Apr 18 11:22:00 2009