Prove $(a+b)^3(b+c)^3(c+d)^3(d+a)^3\ge 16a^2b^2c^2d^2$ Let $a,b,c,d>0$ and such $a+b+c+d=1$, show that
$$(a+b)^3(b+c)^3(c+d)^3(d+a)^3\ge 16a^2b^2c^2d^2$$
since
$$a+b\ge 2\sqrt{ab}$$
I think this will not hold.because we have $256abcd\le 1$

We need to prove that $\prod\limits_{cyc}(a+b)^3\geq16(a+b+c+d)^4a^2b^2c^2d^2$.
Let $a+b+c+d=4u$, $ab+ac+ad+bc+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.
Since $a$, $b$, $c$ and $d$ are positive roots of the equation
$x^4-4ux^3+6v^2x^2-4w^3x+t^4=0$, we see that (by the Rolle's theorem) the equations
$v^2x^2-2w^3x+t^4=0$ and $ux^2-2v^2x+w^3=0$ have two positive roots,
which gives $w^6\geq v^2t^4$ and $v^4\geq uw^3$.
Since $(a+b)(b+c)(c+d)(d+a)-(a+b+c+d)(abc+abd+acd+bcd)=(ac-bd)^2\geq0$, 
we obtain
$\prod\limits_{cyc}(a+b)^3\geq\left(\sum\limits_{cyc}abc\right)^3(a+b+c+d)^3=4^6w^9u^3=\frac{4^6w^{12}u^3}{w^3}\geq\frac{4^6v^4t^8u^3}{w^3}\geq\frac{4^6uw^3t^8u^3}{w^3}=4^6t^8u^4=$
$=16(a+b+c+d)^4a^2b^2c^2d^2$. Done!
