A hard inequality indian olympiad problem If $x,y,z$ are positive real numbers, prove that:
$\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\le 3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)$.
I have been stuck in it. It is an Indian Olympiad problem. Can you guys help me out, please?

We have that:
$$x^2+xy+y^2=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\geq \frac{3}{4}(x+y)^2$$
Therefore
$$(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)\geq \frac{27}{64}(x+y)^2(y+z)^2(z+x)^2$$
And it remains to prove
$$9(x+y)(y+z)(z+x)\geq 8(x+y+z)(xy+yz+zx)$$
We can prove this with AM-GM:
$$8(x+y+z)(xy+yz+zx)=8(x+y)(y+z)(z+x)+8xyz \leq 9(x+y)(y+z)(z+x)$$
