Prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)$

This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).

If $x, y, z$ are positive real numbers, prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$

My best idea was to expand this and simplify. Although that doesn't look very feasible. Another idea is to see that $x^2+y^2+xy \geq x^2+y^2$. Then we just have to show that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(x^2+y^2)(x^2+z^2)(y^2+z^2)$ if that is even true.

Solution 1:

$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz \\ \le x^2+xy+y^2+y^2+yz+z^2+x^2+xz+z^2$

$3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)-(x^2+xy+y^2+y^2+yz+z^2+x^2+xz+z^2)(yz+zx+xy)^2=(x+y+z)^2(y^2z^2-xyz^2+x^2z^2-xy^2z-x^2yz+x^2y^2) \ge0$

Solution 2:

Let $AF=x\ , BF=y\ , CF=z$.

$F-$ Fermat point of $\triangle ABC\ $

So inequality we can rewrite as : $x+y+z\le 3R$ , which is obviously true)