<s>[INST] 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.
 [/INST] WLOG, let $x$ be the smallest, $y= a+x$ and $z = b+x$ for $a, b \ge 0$.  Then we have
$$4(x + y + z)^3 - 27(x^2y + y^2z + z^2x) = 27x^3+27(a+b)x^2+9(a+b)^2x + (a-2b)^2 (4a+b)>0$$
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