Ta có $x^3+y^3-xy(x+y) $$= (x+y)(x^2-xy+y^2)-xy(x+y)$ $=(x+y)(x^2-2xy+y^2) $$=(x+y)(x-y)^2 $$Do x>0,y>0 =>x+y>0 mà $(x-y)^2 \geq 0$ =>$(x+y)(x-y)^2 \geq 0$ =>$x^3+y^3-xy(x+y) \geq 0 $=>$x^3+y^3\geq xy(x+y)$ =>$x^3+y^3+1\geq xy (x+y)+1= xy(x+y)+xyz=xy(x+y+z)$=>$1/( x^3+y^3+1)\leq 1/ [xy(x+y+z) ] $cm $1/( y^3+z^3+1)\leq 1/ [yz(x+y+z) ] $$1/( z^3+x^3+1)\leq 1/ [zx(x+y+z) ]$=>$1/( x^3+y^3+1)+ 1/( y^3+z^3+1)+ 1/( z^3+x^3+1)\leq 1/ [xy(x+y+z)]+ 1/[yz(x+y+z)] +1/ [zx(x+y+z)]$ = $1/xyz=1$
Ta có $x^3+y^3-xy(x+y) $$= (x+y)(x^2-xy+y^2)-xy(x+y)$ $=(x+y)(x^2-2xy+y^2) $$=(x+y)(x-y)^2 $$Do x>0,y>0 =>x+y>0 mà $(x-y)^2 \geq 0$ =>$(x+y)(x-y)^2 \geq 0$ =>$x^3+y^3-xy(x+y) \geq 0 $=>$x^3+y^3\geq xy(x+y)$ =>$x^3+y^3+1\geq xy (x+y)+1= xy(x+y)+xyz=xy(x+y+z)$=>1/( x^3+y^3+1)=<1/ [xy(x+y+z) ] cm $1/( y^3+z^3+1)\leq 1/ [yz(x+y+z) ] $$1/( z^3+x^3+1)\leq 1/ [zx(x+y+z) ]$=>$1/( x^3+y^3+1)+ 1/( y^3+z^3+1)+ 1/( z^3+x^3+1)\leq 1/ [xy(x+y+z)]+ 1/[yz(x+y+z)] +1/ [zx(x+y+z)]$ = $1/xyz=1$
Ta có $x^3+y^3-xy(x+y) $$= (x+y)(x^2-xy+y^2)-xy(x+y)$ $=(x+y)(x^2-2xy+y^2) $$=(x+y)(x-y)^2 $$Do x>0,y>0 =>x+y>0 mà $(x-y)^2 \geq 0$ =>$(x+y)(x-y)^2 \geq 0$ =>$x^3+y^3-xy(x+y) \geq 0 $=>$x^3+y^3\geq xy(x+y)$ =>$x^3+y^3+1\geq xy (x+y)+1= xy(x+y)+xyz=xy(x+y+z)$=>
$1/( x^3+y^3+1)
\l
eq 1/ [xy(x+y+z) ]
$cm $1/( y^3+z^3+1)\leq 1/ [yz(x+y+z) ] $$1/( z^3+x^3+1)\leq 1/ [zx(x+y+z) ]$=>$1/( x^3+y^3+1)+ 1/( y^3+z^3+1)+ 1/( z^3+x^3+1)\leq 1/ [xy(x+y+z)]+ 1/[yz(x+y+z)] +1/ [zx(x+y+z)]$ = $1/xyz=1$