`#huy`
`P=1/x+1/y^2+1/z^3`
`=1/(x^2+y^2)+2(1/(xy)+2xy+2008)(x>0,y>0,x+y<=1)`
Ta có:`x>0,y>0`
`x+y>=2\sqrt{xy}`
`<=>(x+y)^2>=4xy`
`<=>1/x+1/y>=4/(x+y)`
Dấu "=" xãy ra khi `x=y`
Ta có:
`1/(x^2+y^2+2/(xy)+4xy`
`=(1/(x^2+y^2)+1/(2xy))+(4xy+1/(4xy))+5/(4xy)`
`=4/(x^2+2xy+y^2)+2\sqrt{4xy . 1/(4xy)}+5/((x+y)^2)`
`=4/((x+y)^2)+2+5/((x+y)^2)`
`=11/((x+y)^2)>=11`
`=>P>=2027,P_min=2027<=>x=y=1/2`
