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.

As another approach, you could prove it as follows.
Due to Hölder's inequality, we have:
$$
\left(\sum_{cyc} \frac{x^3}{(1+y)(1+z)}\right)\cdot\left(\sum_{cyc} (1+y)\right)\cdot\left(\sum_{cyc} (1+z)\right)\ge(x+y+z)^3\iff \sum_{cyc} \frac{x^3}{(1+y)(1+z)}≥\frac{(x+y+z)^3}{\left(\sum_{cyc} (1+y)\right)\cdot\left(\sum_{cyc} (1+z)\right)}=\frac{(x+y+z)^3}{\left(3+x+y+z\right)^2}
$$
So it remains to prove that:
$$
\frac{(x+y+z)^3}{\left(3+x+y+z\right)^2}≥\frac34
$$
Setting $x+y+z=a$, this is equivalent to:
$$
\frac{a^3}{\left(3+a\right)^2}≥\frac34\iff 4a^3≥27+18a+3a^2\iff 4a^3-3a^2-18a-27≥0
$$
We have equality at $a=3$, so we can factor out $a-3$:
$$
(a-3)\left(4a^2+9a+9\right)≥0
$$
But since
$$
a=x+y+z≥3(xyz)^{\frac{1}{3}}=3\iff a-3≥0
$$
due to AM-GM, this is always true, so we're done.
