Inequalities: BMO 2009/10 Round 2

Question

BMO 2009/10 Round 2 Q.4 asks

Prove that for all positive reals $x,y, z$: $$4(x+y+z)^3 >27(x^2y + y^2z + z^2x)$$

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.

Answer 1

Another way.

Let $\{x,y,z\}=\{a,b,c\}$ where $a\geq b\geq c$.

Thus, by Rearrangement and AM-GM we obtain: $$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$ $$=b(a^2+ac+c^2)\leq b(a+c)^2=4b\left(\frac{a+c}{2}\right)^2\leq4\left(\frac{b+2\cdot\frac{a+c}{2}}{3}\right)^3=\frac{4(x+y+z)^3}{27}.$$