Proof of A.M. $\geq$ G.M. using integration

We like to prove :

Observe that for any  x > 0 :

    

because, so long as the integrand’s  t  lies strictly increasing or decreasing between  x  and  G , the sign  G – x  and of  1/t – 1/G , must be the same.

Equality sign holds iff  x = G.

Now, replace  x  by    and sum up over  i = 1, 2, … , n.

\ A ³ G

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