利用柯西不等式[X^4/(2+y^2-z) +y^4/(2+z^2-x) +z^4/(2+x^2-y)] (2+y^2-z +2+z^2-x +2+x^2-y)>=(x^2+y^2+z^2) 显然x=y=z时等号成立由于x+y+z=1,带入第二个括号(2+y^2-z +2+z^2-x +2+x^2-y),得[X^4/(2+y^2-z) +y^4/(2...