BMO 2009/10 Round 2 Q.4 asks     Prove that for all positive reals $x,y, z$: $$4(x+y+z)^3 &gt;27(x^2y + y^2z + z^2x)$$   My try:-  Using AM-GM, LHS is greater than $108xyz$. Using AM-GM, RHS is greater than $81xyz$. But I cannot think of how to proceed. Please help. Thank you.
Prove that, for all positive real numbers $x$, $y$ and $z$, $$4(x + y + z)^3 &gt; 27(x^2y + y^2z + z^2x)$$  I've tried expressing it as a sum of squares, but haven't got anywhere.   Hints are also welcome.
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