How prove this inequality generalized from 1969 IMO problem 6 
Let 
  $x_{1},x_{2},y_{1},y_{2},z_{1},z_{2},w_{1},w_{2} $ are all positive numbers, and such
  $$x_{1}y_{1}z_{1}-w^3_{1}>0,\; \text{ and }\;x_{2}y_{2}z_{2}-w^3_{2}>0.$$
  show that
  $$\dfrac{16}{(x_{1}+x_{2})(y_{1}+y_{2})(z_{1}+z_{2})-(w_{1}+w_{2})^3}\le\dfrac{1}{x_{1}y_{1}z_{1}-w^3_{1}}+\dfrac{1}{x_{2}y_{2}z_{2}-w^3_{2}}.$$

This problem is created by me, and the background is the 1969 IMO problem 6,
please see: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=363659&sid=efafc24957d8919546afe4254638aea6#p363659
Thank you everyone for showing a proof, I still can't prove it.
 A: It seems the following. 
It is a partial answer, clarifying the situation (for me :-)).
At first we consider the set $D=\{(x,y,z,w)\in (0;\infty)^4: xyz>w^3\}$. We claim that the set $D$ is convex. To show this it suffice to check that the function $g(x,y,z)=–(xyz)^{1/3}$ is convex. 
The Hessian matrix 
$$H(g)=\frac 19\left\|
\begin{matrix}
2x^{-5/3}y^{1/3}z^{1/3} & -x^{-2/3}y^{-2/3}z^{1/3} & -x^{-2/3}y^{1/3}z^{-2/3}\\
-x^{-2/3}y^{-2/3}z^{1/3} & 2x^{1/3}y^{-5/3}z^{1/3} & -x^{-2/3}y^{-2/3}z^{1/3}\\
-x^{-2/3}y^{1/3}z^{-2/3} & -x^{-2/3}y^{-2/3}z^{1/3} & 2x^{1/3}y^{1/3}z^{-5/3}\\
\end{matrix}
\right\|$$
has the following principal minors 
$$\Delta_1=2x^{-5/3}y^{1/3}z^{1/3}\ge 0, \Delta_2=2x^{1/3}y^{-5/3}z^{1/3}\ge 0,
\Delta_3=2x^{1/3}y^{1/3}z^{-5/3}\ge 0,$$
$$\Delta_{12}=3x^{-4/3}y^{-4/3}z^{2/3}\ge 0, \Delta_{13}=3x^{-4/3}y^{2/3}z^{-4/3}\ge 0, 
\Delta_{23}=3x^{1/3}y^{-4/3}z^{-4/3}\ge 0,$$ 
$$\Delta_{123}=0\ge 0.$$  
Since all of them are non-negative, the matrix $H(g)$ is positive semidefinite , so the function $g$ is convex.
Hence the question inequality is equivalent to the convexity of the function $f(x,y,z,w)=\frac 1{xyz-w^3}$ defined on the convex set $D$, which again can be checked as positive semidefiniteness of the Hessian matrix $H(f)$. :-) But I had not done this check, so I don’t know the question inequality holds or not. 
A: use the rearrangement inequality !
firstly, we assume that :
$x_{1}y_{1}z_{1}-w^3_{1}=a>0\; \text{ and }\;x_{2}y_{2}z_{2}-w^3_{2}=b>0$
by observing your inequality , you can find that :
$\dfrac{16}{(x_{1}+x_{2})(y_{1}+y_{2})(z_{1}+z_{2})-(w_{1}+w_{2})^3}\le\dfrac{1}{x_{1}y_{1}z_{1}-w^3_{1}}+\dfrac{1}{x_{2}y_{2}z_{2}-w^3_{2}}$$\Longrightarrow$
$\frac{16}{x_{2}y_{2}z_{1}+x_{1}y_{2}z_{1}+x_{2}y_{1}z_{1}+x_{1}y_{1}z_{2}+x_{1}y_{2}z_{2}+x_{2}y_{1}z_{2}+a+b-3w_{1}w_{2}(w_{1}+w_{2})}$$\le$$\frac{a+b}{ab}$
then, you can use the rearrangement inequality to get :
$4(a+b)^{2}\ge16ab$
the hint is :
firstly, we assume that :
$\frac{x_{1}y_{1}z_{1}}{w_{1}^3}\ge{1}$$,$$\frac{x_{2}y_{2}z_{2}}{w_{2}^3}\ge{1}$
where, $\frac{x_{1}}{w_{1}}$$\ge$$\frac{x_{2}}{w_{2}}$$,$
$\frac{y_{1}}{w_{1}}$$\ge$$\frac{y_{2}}{w_{2}}$$,$
$\frac{z_{1}}{w_{1}}$$\ge$$\frac{z_{2}}{w_{2}}$
if $a<b$, then we can use the rearrangement inequality to transfer :
$x_{2}y_{2}z_{1}+x_{1}y_{2}z_{1}+x_{2}y_{1}z_{1}+x_{1}y_{1}z_{2}+x_{1}y_{2}z_{2}+x_{2}y_{1}z_{2}$
to :
$3(x_{1}y_{1}z_{1}+x_{2}y_{2}z_{2})$
we also transfer :
$3w_{1}w_{2}(w_{1}+w_{2})$
to :
$3w_{1}^{3}+3w_{2}^{3}$
hence, your solution holds !
