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Let $x,y,z$ be positive real numbers. We have to prove that:

$$3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)\geq (x+y+z)^2(xy+yz+zx)^2.$$

Thanks :)

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up vote 1 down vote accepted

Note that $x^2+xy+y^2=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\ge \frac{3}{4}(x+y)^2 $

Repeating the argument for y and z and then for x and z,we get $$3(x^2+y^2+z^2)(y^2+yz+z^2)(x^2+xz+z^2)\ge \frac{81}{64}(x+y)^2(y+z)^2(z+x)^2$$

or $$8(x+y+z)(xy+yz+zx)\leq 9(x+y)(y+z)(z+x)\dots (1)$$ iff $\sum x(y-z)^2\ge 0$ which is true! where the sum is taken cyclically over x,y and z.

P.S. Alternatively,you can proceed from (1) as follows:

$(x+y)(y+z)(z+x)=(x+y+z)(xy+xz+yz)-xyz$ which reduces the problem to proving $(x+y)(y+z)(z+x)\ge 8xyz$ which is true by AM-GM.

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It is Indian NMO 2007, Problem 6. See here for lots of different solutions.

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