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?
 A: Let $p = x+y+z, \ q = xy+yz+zx, \ r = xyz$. 
Rewrite the inequality as
$p^2 q^2 \le 3(p^2q^2 - q^3 - p^3 r)$.
Since $q^2 \ge 3pr$, it suffices to prove that
$$p^2 q^2 \le 3(p^2q^2 - q^3 - p^3 \frac{q^2}{3p})$$
or
$$q^2(p^2-3q)\ge 0.$$
It is obvious. We are done.
A: 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:
$$8(x+y+z)(xy+yz+zx)=8(x+y)(y+z)(z+x)+8xyz \leq 9(x+y)(y+z)(z+x)$$
A: $\quad3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)-\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\\=\sum_{sym} \left(2x^4y^2z^0+0.5x^3y^3z^0+0.5x^4y^1z^1-2x^3y^2z^1-x^2y^2z^2\right)\\=2\sum_{sym} \left(x^4y^2z^0-x^3y^2z^1\right)+0.5\sum_{sym} \left(x^3y^3z^0-x^2y^2z^2\right)+0.5\sum_{sym} \left(x^4y^1z^1-x^2y^2z^2\right)$
By Muirhead's Inequality, $\left(4,2,0\right)\succ\left(3,2,1\right),\left(3,3,0\right)\succ\left(2,2,2\right),\left(4,1,1\right)\succ\left(2,2,2\right)$
$\because$ The expression $\ge 0$, equality holds when $x=y=z$
$\therefore 3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)\ge\left(x+y+z\right)^2\left(yz+xz+xy\right)^2$
($\succ$ denotes majorization)
