parmenides51 wrote: Determine the smallest value of $x^2 + y^2 + z^2$, where $x, y, z$ are real numbers, so that $x^3 + y^3 + z^3 -3xyz = 1.$ Let $x, y, z$ are real numbers such that $x^3 + y^3 + z^3 -3xyz = 1.$ Prove that$$x^2 + y^2 + z^2\geq 1$$

We will prove that $(x^2+y^2+z^2)^3 \ge (x^3+y^3+z^3-3xyz)^2$. Let $p = x+y+z , q = xy+yz+zx$. It's equivalent to $(p^2-2q)^3 \ge (p^3-3pq)^2$ which after expanding simplifies to $q^2(3p^2-8q)\ge 0$ which is true since $a^2+b^2+c^2 + (a-b)^2+(b-c)^2+(c-a)^2 \ge 0$. equality occurs when $xy+yz+xz = q = 0$.