Three Variables Inequality $a$,$b$,$c$ are non-negative numbers such that $(a+b)(b+c)(c+a)=1$.    
show that $\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)}\geq\sqrt[3]{2}\sqrt{ab+bc+ca}$    
I can't guess where the constant $\sqrt[3]{2}$ come from.

let $ x=b+c,y=a+c,z=a+b,\implies xyz=1,a = min${$a,b,c$}$,a=\dfrac{y+z-x}{2} \ge 0 \implies x \le y+z$
$1=xyz \le x\left(\dfrac{y+z}{2}\right)^2 \le \dfrac{(y+z)^3}{4} \implies y+z \ge \sqrt[3]{4} \implies 2a+b+c \ge \sqrt[3]{4} $ 
RHS$=\sqrt{\sqrt[3]{4}(ab+bc+ca)} \le \sqrt{(2a+b+c)(ab+bc+ca)}=\sqrt{bc(b+c)+(2a^2b+2a^2c+4abc+ab^2+ac^2)} $  
$\sqrt{ab(a+b)}+\sqrt{ca(c+a)}\ge \sqrt{2(ab(a+b)+ca(c+a))}  \iff \sqrt{2(ab(a+b)+ca(c+a))} +\sqrt{bc(b+c)} \ge \sqrt{bc(b+c)+(2a^2b+2a^2c+4abc+ab^2+ac^2)}\iff  ab^2+ac^2 +2\sqrt{2(ab(a+b)+ca(c+a))bc(b+c)} \ge 4abc$
$ab^2+ac^2 \ge 2abc \iff \sqrt{2(ab(a+b)+ca(c+a))bc(b+c)} \ge abc \iff abc(2b^3+2ac^3+2ab^2+2ac^2+3abc) \ge 0$
all above "=" hold when $a=0,b=c$ must be hold both or either. so the inequality will be hold "=" when $a=0$ and $b=c$
QED.
