Let $x+y+z=1, x\geq 0,y\geq 0,z\geq 0$. Prove that $$ x^2y + y^2z + z^2x \leq 4/27 $$
I know how to prove it by setting $z=0$ and finishing off with Lagrange multipliers with the new constraint $x+y=1$:
Let $f(x,y,z) = x^2y + y^2z + z^2y$ and assume w.l.o.g. that $x\geq y$ and $x\geq z$, since $f$ is cyclic. Next, it is easy to prove that $f(x,y,z)\geq f(x,z,y)$, so we can also assume w.l.o.g. that $x\geq y \geq z$. Similarly, $f(x+z,y,0) \geq f(x,y,z)$, so $z=0$ w.l.o.g. Finishing off, using Lagrange multipliers and solving the system \begin{align*} 2xy = \lambda,\\ x^2 = \lambda,\\ x+y = 1, \end{align*} we get the maximizer $(x^*,y^*)=(2/3,1/3)$, and $f(2/3,1/3,0) = 4/27$.
What I want is to prove this in a different way, e.g. by using AM-GM. Some ideas I considered is using AM-GM: $$ x^2y + y^2z + z^2y \leq \frac{(2x+y)^3 +(2y+z)^3+(2z+y)^3}{27}, $$ or proving the equivalent inequality: $$ 27(x^2y + y^2z + z^2x) \leq 4(x+y+z)^3. $$
Edit: while the question has been answered before (and for some reason I didn't see it), I couldn't help but notice that the linked question contains only 2 answers, with one downvoted and similar to my proof. Both of these also make use of $x\geq y \geq z$, which requires some initial justification. So, I'm particularly interested in more proofs, so I can learn to tackle more inequalities of similar nature.
You may finish it off with AM-GM if you want to avoid Lagrange multipliers:
$$x^2y=4\left(\frac{x}{2}\cdot\frac{x}{2}\cdot y\right)\le4\left(\frac{\frac{x}{2}+\frac{x}{2}+y}{3}\right)^3=\frac{4}{27}$$
with equality when $\frac{x}{2}=y=\frac{1}{3}$
Your last inequality can also be proved with BW method, see Prove the inequality $4(a+b+c)^3\ge 27(a^2b+b^2c+c^2a)$
And it has been asked before as well, see Prove that, for all non-negative real numbers $x,y,z$ that satisfy $x+y+z=1, x^2y+y^2z+z^2x≤4/27$
As your proof of $x\ge y\ge z$ has some value not found in the other posts, I'm not voting to close the question, but others might have a different opinion.