Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y, z \ge 0$. 
Given non-negatives $x, y, z$ such that $x + y + z = 4$. Calculate the maximum value of $$\large x^3y + y^3z + z^3x$$

As an assumption, the maximum value is $27$, occured when $(x, y, z) = (0, 1, 3)$.
I have a guess about a working-in-process prove. Let $y$ be the median of $x, y, z$.
$$\iff (zx - yz)(y^2 - z^2) \ge 0 \iff y^2zx - y^3z - z^3x + yz^3 \ge 0$$
$$\iff x^3y + y^3z + z^3x \le x^3y + y^2zx + yz^3 = y(x^3 + xyz + z^3)$$
And further down the line is what I haven't accomplished yet.

I like the following way.
Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Thus, by Rearrangement and AM-GM we obtain:
$$x^3y+y^3z+z^3x=x^2\cdot xy+y^2\cdot yz+z^2\cdot zx\leq a^2\cdot ab+b^2\cdot ac+c^2\cdot bc=$$
$$=b(a^3+c^3+abc)\leq b(a+c)^3=27\cdot b\left(\frac{a+c}{3}\right)^3\leq27\left(\frac{b+3\cdot\frac{a+c}{3}}{4}\right)^4=27.$$
