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.
 A: 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.
A: 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 $x=y=z=1$
A: I think you can do this using Lagrange multipliers. Define the function:
$$f(x,y,z)=  \frac{x^3}{(1+y)(1+z)} +\frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)} - \frac{3}{4}.$$
We want to show that $f(x,y,z)\geq 0$ for all $x,y,z\in \mathbb R^3$ such that $xyz=1$. Define the Lagrangian:
$$L(x,y,z,\lambda) = f(x,y,z)- \lambda g(x,y,z)$$
where $g(x,y,z)=0$ is the function defining the set where you look at: that is $xyz=1$ and hence $g(x,y,z)=xyz-1$. As a result,
$$L(x,y,z,\lambda) = f(x,y,z)- \lambda (xyz-1).$$
Differentiating $L$ w.r.t. all variables we get
\begin{align*}
\partial_x L(x,y,z,\lambda) =& \frac{3x^2}{(1+y)(1+z)} -\frac{y^3}{(1+x)^2(1+z)} - \frac{z^3}{(1+x)^2(1+y)} -\lambda y z = 0\\
\partial_y L(x,y,z,\lambda) =& -\frac{x^3}{(1+y)^2(1+z)} +\frac{3y^2}{(1+x)(1+z)}  -\frac{z^3}{(1+x)(1+y)^2} - \lambda xz=0\\
\partial_z L(x,y,z,\lambda) =& -\frac{x^3}{(1+y)(1+z)^2} -\frac{y^3}{(1+x)(1+z)^2} + \frac{3z^2}{(1+x)(1+y)} - \lambda x y = 0\\
\partial_{\lambda} L(x,y,z,\lambda) =& 1-xyz=0
\end{align*}
Now try to exploit the symmetries or similarities in order to solve this system. The critical points can be maxima, minima or saddle points. If you happen to find only minima and such points satisfy $f(x,y,z)\geq 0$ then you will have proven the inequality. Since you are proving that the function $f$ has minima in the set $\{(x,y,z)\in \mathbb{R}^3 \, xyz=1\}$ and all of them are positive.
Be careful that the function is continuous and the set you are looking at is closed. 
Another possibility is to subsitute the vaue for $z=1/(xy)$ (it is ok since $x,y,z\neq 0$) in the inequality and define a two-variable function $f(x,y)$ and do exactly the same without using Lagrange Method. Maybe that is easier.
