Least value

Using AM-GM on the three terms in the expression, we find that the sum is greater than or equal to three times the cube root of $\frac{x^3y^3z^3}{(x^2+y)(y^2+z)(z^2+x)}$ , which simplifies to $\frac{1}{(x^2+y)(y^2+z)(z^2+x)}$ . Using AM-GM on the denominator, we that the cube root of $(x^2+y)(y^2+z)(z^2+x)$ is less than or equal to $\frac{x^2 + y^2 + z^2 + x + y + z}{3} = 2$ . Thus, taking the reciprocals of both sides (since all variables are positive), we find that $\frac{x^2 + y^2 + z^2 + x + y + z}{3}$ is greater than or equal to 1/2 and that the original sum is greater than or equal to 3/2. Equality is achieved when x=y=z=1, proving that 3/2 is the minimum value.