Let $x,y,z>0,xyz=1$. Prove that $\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\ge \frac34$ Let $x,y,z>0$ and $xyz=1$. Prove that $\dfrac{x^3}{(1+y)(1+z)}+\dfrac{y^3}{(1+x)(1+z)}+\dfrac{z^3}{(1+x)(1+y)}\ge \dfrac34$  
My attempt:  
Since it is given that $xyz=1$, I tried substituting $x=\dfrac{a}{b},y=\dfrac{b}{c},z=\dfrac{c}{a}$. But the expansion looked really ugly and I didn't think I could make out anything out of it.
So, I made another attempt, if each element was greater than $\dfrac{1}{4}$, we could have a solution, so, treating that way, I get $4x^3\ge 1+x+y+xy, 4y^3\ge 1+x+z+xz, 4z^3\ge 1+x+y+xy$. Using AM-GM I get an equality.
So, please help. Thank you.

Edit1:
another approach is to use $\dfrac{x^2}{a}+\dfrac{y^2}{b} \ge \dfrac{(x+y)^2}{a+b},x+y+z\ge3$
LHS $\ge \dfrac{(x^2+y^2+z^2)^2}{\sum_{cyc} {(1+y)(1+z)}{x}}\ge\dfrac{(x+y+z)^4}{9(x+y+z+2(xy+yz+xz)+3)}\ge \dfrac{(x+y+z)^4}{9(x+x+z)+6(x+y+z)^2+27}\ge\dfrac{(x+y+z)^4}{9(x+y+z)+6(x+y+z)^2+9(x+y+z)}=\dfrac{(x+y+z)^3}{18+6(x+y+z)}\ge\dfrac{(x+y+z)^3}{6(x+x+z)+6(x+y+z)}=\dfrac{(x+y+z)^2}{12}\ge \dfrac{3^2}{12}=\dfrac{3}{4} $
all "=" hold when $x=y=z=1$
