Prove $4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$ Prove that, for all positive real numbers $x$, $y$ and $z$,
$$4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$$
I've tried expressing it as a sum of squares, but haven't got anywhere. 
Hints are also welcome.

This is well-know inequality, actually this is a inequality from Vacs if you are familiar with Art of Problem Solving. Here's the proof:
Let $z\le y \le x$, so we have:
$$(y-x)(y-z) \le 0 \implies y^2 - xy - zy + xz \le 0 \implies y^2 + nz \le xy + zy$$
Now for the Right Hand Side:
$$27(x^2y + y^2z + z^2x + xyz) = 27( z(y^2 + zx) + x^2y + xyz) \le 27 (z(xy + zy) + x^2y + xyz) = 27 ( 2xyz + z^2y + x^2y) = 27y(2xz + x^2 + z^2) = 27y(x+z)^2 = 4 \cdot 3^3y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right)$$
Now from AM-GM we have:
$$y + \frac{x+z}{2} + \frac{x+z}{2} \ge 3 \sqrt{y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right)}$$
$$(y + \frac{x+z}{2} + \frac{x+z}{2})^3 \ge 3^3 y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right)$$
So we have:
$$27(x^2y + y^2z + z^x + xyz) \le 4 \cdot 3^3y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right) \le (y + \frac{x+z}{2} + \frac{x+z}{2})^3 = 4(x+ y+z)^3$$
Hence the proof. As you can note this inequality is even stronger, because $xyz > 0$
