Áp dụng bất đẳng thức AM-GM, ta có:

`2x^2 + y^2 + 3 = (x^2 + y^2) + (x^2 + 1) + 2 >= 2xy + 2x + 2 = 2(xy + x + 1)`

`=> x / (2x^2 + y^2 + 3) <= x / (2(xy + x + 1))`

Tương tự, ta có:

`y / (2y^2 + z^2 + 3) <= y / (2(yz + y + 1))`

`z / (2z^2 + x^2 + 3) <= z / (2(zx + z + 1))`

`=> P <= 1/2 [x / (xy + x + 1) + y / (yz + y + 1) + z / (zx + z + 1)]`

Ta lại có: `x / (xy + x + 1) + y / (yz + y + 1) + z / (zx + z + 1)`

`= x / (xy + x + 1) + (xy) / (xyz + xy + x) + (xyz) / (x^2yz + xyz + xy)`

`= x / (xy + x + 1) + (xy) / (1 + xy + x) + 1 / (x + 1 + xy)`

`= (x + xy + 1) / (xy + x + 1) = 1`

Suy ra: `P <= 1/2 * 1 = 1/2`

Dấu "`=`" xảy ra khi `x = y = z = 1`

Vậy `P_{max} = 1/2` khi `x =y = z = 1`