<s>[INST] 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?
 [/INST] $\quad3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)-\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\\=\sum_{sym} \left(2x^4y^2z^0+0.5x^3y^3z^0+0.5x^4y^1z^1-2x^3y^2z^1-x^2y^2z^2\right)\\=2\sum_{sym} \left(x^4y^2z^0-x^3y^2z^1\right)+0.5\sum_{sym} \left(x^3y^3z^0-x^2y^2z^2\right)+0.5\sum_{sym} \left(x^4y^1z^1-x^2y^2z^2\right)$
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)
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