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.

Now, use AM-GM:
$$y(x^3+xyz+z^3)\leq y(x+z)^3=27y\left(\frac{x+z}{3}\right)^3\leq27\left(\frac{y+3\cdot\frac{x+z}{3}}{4}\right)^4=27.$$
