proof of AM-GM

Rearrangement inequality:
if a1<= a2 <= a3 and b1<=b2<=b3, then
a1b1+a2b2+a3b3>= a1b2+a2b3+a3b1>=a1b3+a2b2+a3b1 -----(*)
(Notice the order of their inner product.)

Without loss of generality, we can suppose that 0<=a<=b<=c, then
(1) a<=b<=c
(2) a^2 <= b^2 <= c^2
(3) ab <= ac <= bc
(A) From (1),(2) and by rearrangement inequality(left part of (*)), then
a^3+b^3+c^3 >= a^2b+b^2c+c^2a=(ab)a+(ac)c+(bc)b ----(**)
(B) From (3),(1) and the right part of (*), then
(ab)a+(ac)c+(bc)b>=(ab)c+(ac)b+(bc)a= 3abc -----(***)
(C) From (**)and(***) then a^3+b^3+c^3 >= 3abc
(D) Let a^3=x, b^3=y, c^3=z, then a^3+b^3+c^3>= 3abc will become
x+y+z >= 3 (xyz)^(1/3) or (x+y+z)/3 >= (xyz)^(1/3)
ie. AM>=GM