How to prove this inequality without using Muirhead's inequality? I ran into a following problem in The Cauchy-Schwarz Master Class:

Let $x, y, z \geq 0$ and $xyz = 1$.
Prove $x^2 + y^2 + z^2 \leq x^3 + y^3 + z^3$.

The problem is contained in the chapter about symmetric polynomials and Muirhead's inequality.
The proof based on Muirhead's inequality is pretty quick:
We multiply the left hand side with $\sqrt[3]{xyz} = 1$ and prove
$$x^{\frac{7}{3}}y^{\frac{1}{3}}z^{\frac{1}{3}} + x^{\frac{1}{3}}y^{\frac{7}{3}}z^{\frac{1}{3}} + x^{\frac{1}{3}}y^{\frac{1}{3}}z^{\frac{7}{3}} \leq x^3 + y^3 + z^3$$
with Muirhead ( $(3, 0, 0)$ majorizes $(\frac{7}{3}, \frac{1}{3}, \frac{1}{3})$).
I'm curious if there's a way to prove this without machinery of Muirhead's inequality and majorization. Also, this approach readily generalizes to proving
$$x^n + y^n + z^n \leq x^{n+1} + y^{n+1} + z^{n+1}$$
for non-negative $x, y, z$ such that $xyz = 1$.
Is there a way to prove this generalization without Muirhead?

Another approach, not using the general power inequalities.
For convenience, we say $a=x^{\frac 13}$ and similar for $b$ and $c$.
Just apply the inequality arithmetic mean $>$ geometric mean in the following way (and cyclic permutations):
\begin{align}
\frac 79 a^9+\frac 19 b^9+\frac 19 c^9&=\frac{\underbrace{a^9+a^9+\cdots+a^9}_{\text{7 times}}+b^9+c^9}{9}\\
&\geq \sqrt[9]{\left(a^9\right)^7 b^9c^9}=a^7bc
\end{align}
Summing the cyclic permutations gives the required inequality.
In the general version, use the following: (where $a=x^{\frac 13}$ again)
\begin{align}
\frac{\underbrace{a^{3n+3}+a^{3n+3}+\cdots+a^{3n+3}}_{\text{3n+1 times}}+b^{3n+3}+c^{3n+3}}{3n+3}
&\geq \sqrt[b^{3n+3}]{\left(a^{3n+3}\right)^{3n+1} b^{3n+3}c^{3n+3}}=a^{3n+1}bc
\end{align}
