Từ gt ta có: `8+4a+4b+4c+2ab+2bc+2ca=1`

`<=> (a+2)(b+2)(c+2) = 1`

Đặt: `x=a+2=>a=x-2`

        `y=b+2=>b=y-2`

        `z=c+2=>c=z-2`

`=> xyz=1 => {(yz=1/x),(z=1/(xy)),(xz=1/y):}`

Xét mẫu số ta có:

`7+3a+2b+ab=3+a+(a+2)(b+2)`

`<=> 7+3a+2b+ab=3+x-2+xy`

`<=> 7+3a+2b+ab=1+x+xy`

Tương tự ta có:

`7+3b+2c+bc=1+y+yz`

`7+3c+2a+ac=1+z+xz`

`=> P = 1/(1+x+xy) + 1/(1+y+yz) + 1/(1+z+xz)`

`<=> P = 1/(1+x+xy) + 1/(1+y+1/x) + 1/(1+1/(xy)+1/y)`

`<=> P = 1/(1+x+xy) + x/(x+xy+1) + (xy)/(xy+1+x)`

`<=> P = (1+x+xy)/(1+x+xy) = 1`

Vậy `P=1`