1999年加拿大数学奥林匹克试题第5题为:已知x,y,z为满足x+y+z=1的非负实数,试证:x~2y+y~2z+z~2x≤4/(27),并指出等号成立的条件.将之加强,则有命题1若x,y,z是非负实数,则x~2y+y~2z+z~2x≤4/(27)(x+y+z)~3(1)证明所证不等式左边是关于x,y,z的轮换对称结构,不妨设y是x,y,z三个数中的中间数, In 1999, the fifth question of Mathematical Olympiad in Canada was: It is known that x, y and z are nonnegative real numbers satisfying x + y + z = 1, and test: x ~ 2y + y ~ 2z + z ~ 2x≤4 / ( 27), and pointed out the conditions for the establishment of the equal sign. To reinforce it, there is a proposition 1 if x, y, z are nonnegative real numbers, then x ~ 2y + y ~ 2z + z ~ 2x≤4 / (27) y + z) ~ 3 (1) Prove that the left side of the certified inequality is about the rotational symmetry structure of x, y and z. Let y be the middle of the three numbers x, y and z,