~~ 超級算術幾何不等式 ~~

|√x-√y| >=|√y-√z|>=|√z-√x|
Without loss of generality, we can assume that x>=z>=y>=0,
and x=a^6 y, z=b^6 y, 1<= b<=a.
√z-√y >=√x-√z, then b³>=(a³+1)/2 >=√a³, so, b>=√a
Hence, a >= b >=√a >= 1
(√x-√y)²-(x+y+z)/3+∛(xyz) >=0 <=> 3(a³-1)²-(a^6+b^6+1)+3a²b² >=0
Consider f(t)=3(a³-1)²-(a^6+t^6+1)+3a²t² for √a <= t (fixed a)
f'(t)= -6t^5+6a²t
=6t^5[(a/t²)²-1] <=0 (since √a <= t)
so, f(b) >= f(a)=3(a³-1)²-(2a^6+1)+3a^4, for a>=1
Let g(t)=3(t³-1)²-(2t^6+1)+3t^4 for t >=1
g'(t)=18t²(t³-1)-12t^5+12t³
=6t²(t³+2t-3) >=0
then g(t)>=g(1)=0.
so, f(b) >= f(a)=g(a) >= g(1) =0

2012-05-12 23:51:29 補充：
Consider f(t)=3(a³-1)²-(a^6+t^6+1)+3a²t² for √a <= t <=a (fixed a)