Sum of squares <= Sum of cubes

Source: http://ruffnsluff.blogspot.com/2020/04/sum-of-squares-sum-of-cubes.html

Suppose $x, y, z>0$ and $xyz=1$. Show that $x^3+y^3+z^3\ge x^2+y^2+z^2$.

My solution is far from elegant.

Proof:

We try to show $f\left(x,y,z\right)=x^3+y^3+z^3-x^2-y^2-z^2\geq 0$. By fixing one variable and taking partial derivative to zero, a necessary condition for such $\left(x,y,z\right)$ is that

$$
3x^3-2x^2=3y^3-2y^2=3z^3-2z^2.
$$

There are three cases.

(i) $x,y,z$ are distinct. Then they are roots of polynomial $3p^3-2p^2-3=0$. It is not hard to find out the corresponding $f\left(x,y,z\right)$, which is positive so not the minimum $0$ we want to show.

(ii) $x=y=z=1$. Trivial.

(iii) $x=y\neq z$. Setting partial derivative to zero gives us

$$
3x^9-2x^8+2x^2-3=0,
$$

which has only one real root $x=1$, i.e. $x=y=z=1$.

Q. E. D.