Có `x^2+y^2=1`
`<=>(x+y)^2+2xy=1`
`=>xy={(x+y)^2-1}/2`
Có `P=(3-x)(3-y)`
`=9-3x-3y-xy`
`=9-3(x+y)+{(x+y)^2}/2-1/2`
`=1/2(x+y)^2-3(x+y)+17/2`
`={[1/{\sqrt{2}}(x+y)]^2-2. 1/{\sqrt{2}}. {3\sqrt{2}}/2(x+y)+9/2}+8/2`
`={1/{\sqrt{2}}(x+y)-{3\sqrt{2}}/2}^2+4`
Có `{1/{\sqrt{2}}(x+y)-{3\sqrt{2}}/2}^2>=0` `∀x,y∈RR`
`=>P>=4`
Vậy $P_{min}$ `=4<=>x+y=3`