試解

去分母得等價不等式：
x(y^2+2)(z^2+2) + y(z^2+2)(x^2+2) + z(x^2+2)(y^2+2) <=
(x^2+2)(y^2+2)(z^2+2)

由xyz=1,
左邊 = (xy+yz+zx) + 4(x+y+z) + 2(xy(x+y) + yz(y+z) + zx(z+x)) =
A + B + C

右邊 = 2(x^2y^2+y^2z^2+z^2x^2) + 4(x^2+y^2+z^2) + 9 >=
(3/2)(x^2y^2+y^2z^2+z^2x^2) + 4(x^2+y^2+z^2) + 9 + 3/2.

由xy <= x^2y^2/2 + 1/2
A <= (1/2)(x^2y^2+y^2z^2+z^2x^2) + 3/2;

由4x <= 2x^2+2
B <= 2(x^2+y^2+z^2) + 6;

由2xy(x+y) <= (x+y)^2 <= x^2+y^2+x^2y^2+1
C <= 2(x^2+y^2+z^2) + (x^2y^2+y^2z^2+z^2x^2) + 3.