Prove that if $a,b,c$ are positive real numbers and $a+b+c = 1$ then $(3abc)^2 \ge (1-2a)(1-2b)(1-2c)(a^2+b^2+c^2).$ 
Prove that if $a,b,c$ are positive real numbers and $a+b+c = 1$ then $(3abc)^2 \ge (1-2a)(1-2b)(1-2c)(a^2+b^2+c^2).$ 

Cauchy-Schwarz doesn't really help here and neither does rearrangement so I see nothing better to do than substitute: $$(1-2a)(1-2b)(1-2c)(a^2+b^2+c^2) = (2(b+c)-1)(2(a+c)-1)(2(a+b)-1)(a^2+b^2+c^2).$$ I get stuck here.

We need to prove that 
$$9a^2b^2c^2\geq(a^2+b^2+c^2)(a+b-c)(a+c-b)(b+c-a)(a+b+c)$$
Let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Since $\prod\limits_{cyc}(a+b-c)>0$, we can assume that $x$, $y$ and $z$ are non-negatives.
Indeed, let $x<0$ and $y<0$. Hence, $x+y<0$, which gives $c<0$. Contradiction.
Thus, we need to prove that
$$9(x+y)^2(x+z)^2(y+z)^2\geq32xyz(x+y+z)\sum\limits_{cyc}(x^2+xy)$$
which is $\sum\limits_{sym}(9x^4y^2+9x^3y^3-7x^4yz-10x^3y^3z-x^2y^2z^2)\geq0$,
which is obvious by Muirhead. 
