For $x+y+z=3$ prove that $4\geqq x^2y+y^2z+z^2x$ Given $3$ positive real numbers $x,\,y,\,z$ and $x+ y+ z= 3$. Prove that:
$$4\geqslant x^2y+ y^2z+ z^2x$$
This problem was homogenized so I set $x+ y+ z= 3$ to cancel stuff.
Now I'm stuck. I have noticed a funny equality condition:
If $(\,x,\,y,\,z\,)=(\, 2,\,1,\,0\,)$ then we have equality we don't have equality if they are all equal.
To me$,$ that implied a weighted AM-GM but I couldn't make that work.
Thanks in advance for any help!
 A: Assume to have a fixed $z\in(0,3]$, then maximize $f(x,y)=x^2 y + y^2 z + z^2 x$ under the constraints $x+y=3-z$ and $x,y>0$. This is the same as maximizing
$$ g(x)=-x^3 + 3 x^2 + (3 z^2 - 6 z) x + (9 z + z^3) $$
under the only constraint $x\in(0,3-z)$. Since:
$$ g'(x) = -3((x-1)^2-(z-1)^2)=3(z-x)(x+z-2)$$
the stationary points of $g(x)$ occur when $x=z$ or $x+z=2$. This gives that it is sufficient to prove the starting inequality when $(x,y,z)=(z,3-2z,z)$ with $z\in(0,3/2)$ and when $(x,y,z)=(2-z,1,z)$ with $z\in(0,2)$. In the first case, we have to prove:
$$ z\in(0,3/2)\quad\Longrightarrow\quad 3z^3-9z^2+9z\leq 4$$
that is straightforward since $\frac{d}{dz}(3z^3-9z^2+9z)=3(z-1)^2\geq 0$ and the inequality holds for $z=\frac{3}{2}$. In the latter case, we have to prove:
$$ z\in(0,2)\quad\Longrightarrow\quad z(z^2-3z+3)\geq 0$$
that is also straightforward.
A: Use this well known 
$$4(\,x+ y+ z\,)^{\,3}- 27(\,xy^{\,2}+ yz^{\,2}+ zx^{\,2}+ xyz\,)\geqq 0$$
A: this is not a solution to your problem, but may shed a little light on the fact you found puzzling
constrained optimization finds the only critical point is $(1,1,1)$. this is a minimum, so the maximum occurs on the boundary.
$$
\frac{\partial}{\partial x}\left(x^2y+y^2z+z^2x -\lambda(x+y+z) \right) =0
$$
gives:
$$
2xy + z^2 = \lambda
$$
subtracting this from the two similar equations for $y,z$ gives $x=y=z$
now, setting one of the variables, say $y$ to zero, for the boundary, gives, by a similar procedure: 
$$
2zx=\lambda = z^2
$$
one of whose two solutions gives the maximum you noticed at $(1,0,2)$.
A: It suffices to show that
$$4(\,x+ y+ z\,)^{\,3}- 27(\,x^{\,2} y+ y^{\,2}z+ z^{\,2}x+ xyz\,)\geqslant 0$$
Can let
$$F= 4(\,x+ y+ z\,)^{\,3}- 27(\,x^{\,2} y+ y^{\,2}z+ z^{\,2}x+ xyz\,)$$
$$\therefore F= (\,x+ 4\,z\,)(\,y+ z- 2\,x\,)^{\,2}+ 4\,y(\,y- z\,)^{\,2}- 11y(\,x- y\,)(\,x- z\,)\geqslant  0$$
$$\because\,x\equiv \text{mid}\{x,\,y,\,z\} \tag{assume}$$
A: Using Ji Chen 's decomposition
Assuming $x= \min\{\,x,\,y,\,z\,\}$
$$\therefore\,4(\,x+ y+ z\,)^{\,3}- 27(\,xy^{\,2}+ yz^{\,2}+ zx^{\,2}\,) $$
$$= (\,x+y-2\,z\,)^{\,2}(4\,y+ \,z- 5\,x\,)+ 9x\,(\,x^{\,2}+ y^{\,2}+ z^{\,2}- xy- yz-\,zx\,)\geqq 0$$
