Avoiding multiplication in inequality

Question

For $x,y,z\in \mathbb R^{+}$, prove that: $$(x+y+z)^2(xy+yz+zx)^2\le3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+xz+x^2)$$ I have been trying to attack this problem by setting $a=x^2+xy+y^2, b=y^2+yz+z^2, c=y^2+yz+z^2$ and searching for those terms in the LHS using: $$3(xy+yz+zx)\le 2(x^2+y^2+z^2)+xy+yz+zx=a+b+c$$ $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\le a+b+c$$ But I think there is no hope in this method since I would be left with a $3abc$ in the RHS, which is almost never greater than something. But I haven't come up with another way to avoid multiplying the RHS(since using $x^2+y^2\ge2xy$ is not productive). Another idea is to use some sort of geometric argument by setting lengths $x,y,z$ with a $120°$ angle between them, but it hasn't worked so far. Any help would be greatly appreciated.

Answer 1

Your geometry idea is nice, infact it works :) . Consider a triangle with sides $ABC$ in which there is a point $H$ such that $AH=x, BH=y, CH=z$ and $\angle AHB = \angle BHC = \angle CHA = 120^{\circ}$. Then using cosine, rule gives $a^2= y^2+z^2+yz , b^2=x^2+z^2+xz , c^2 = x^2+y^2+xy $. Also since sum of area of these three triangles, equals area of $ABC$ this implies that, $xy+yz+xz= \frac{4A}{\sqrt{3}}$. Now, $(x+y+z)^2= \frac{a^2+b^2+c^2+4A\sqrt{3}}{2} $. So given inequality is equivalent to, $$ \frac{a^2+b^2+c^2+4A\sqrt{3}}{2} \times \frac{16A^2}{3} \le 3a^2b^2c^2 $$ Writing, $abc = 4AR$ , this inequality is equivalent to, $$a^2+b^2+c^2+4\sqrt{3}A \le 18R^2 $$ This is true, because distance between circumcentre of triangle and orthocentre of triangle is precisely $\sqrt{9R^2 -a^2-b^2-c^2}$ . And so, $9R^2 \ge a^2+b^2+c^2 $. And it is very well known that $a^2+b^2+c^2\ge 4\sqrt{3}A $ $\Box$