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.
 A: 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 $xyz > 0$
A: If $x,y,z>0$, then $$x(2x-4y-z)^2+y(2y-4z-x)^2+z(2z-4x-y)^2\gt 0$$
Rearranging gives the required inequality.
A: Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Thus, by Rearrangement and AM-GM we obtain:
$$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$
$$=a^2b+abc+c^2b\leq a^2b+2abc+c^2b=b(a+c)^2=4\cdot b\left(\frac{a+c}{2}\right)^2\leq$$
$$\leq4\left(\frac{b+2\cdot\frac{a+c}{2}}{3}\right)^3=\frac{4}{27}(a+b+c)^3=\frac{4}{27}(x+y+z)^3.$$
A: WLOG, let $x$ be the smallest, $y= a+x$ and $z = b+x$ for $a, b \ge 0$.  Then we have
$$4(x + y + z)^3 - 27(x^2y + y^2z + z^2x) = 27x^3+27(a+b)x^2+9(a+b)^2x + (a-2b)^2 (4a+b)>0$$
