How find this inequality is $a^3+b^3+c^3+3xabc+y(a+b+c)\ge z(ab+bc+ac)$ 
let $x,y,z\ge 0$ ,Assmue that 
  $$a^3+b^3+c^3+3xabc+y(a+b+c)\ge z(ab+bc+ac),\forall a,b,c\ge 0$$ if and only if
  $$z^2\le \min{(16y,4y(1+x))}$$

My idea： $$\Longrightarrow $$
let $a=b=0,c>0$,then we have
$$c^3+yc\ge 0\Longrightarrow c(c^2+y)\ge 0$$
let 
$a=b>0,c=0$, then we have
$$2a^3++2ay\ge za^2\Longrightarrow 2a^2+2y\ge za$$
then I can't,
Thank you

The necessary condition:
$a=b=c$ gives $a^2(1+x)-za+y \ge 0, \: \forall a \in \mathbb R \implies z^2 \le 4(1+x)y$
Further, $a=b, c = 0$ gives $2a^2-za+2y \ge 0, \: \forall a \in \mathbb R \implies z^2 \le 16y$, so together, we have $$z^2 \le 4y\;\min(4, 1+x).$$  
The sufficient condition:
An easy way would be to use the symmetric function theorem$^\dagger$, i.e. for a symmetric polynomial of degree three, $P(x_1, x_2, \dots, x_n)$, we have
$$\forall x_i \ge 0, \;\; P(x_1, x_2, \dots, x_n) \ge 0 \iff \forall k \in \{1,2,\dots,n \},\;\; P(\underbrace{1, 1,\dots 1}_{k \text{ ones}}, 0, ...0) \ge 0$$
So in the case of our three variable cubic, we only need to show $P(1, 0, 0), P(1, 1, 0), P(1, 1,  1) \ge 0$.  i.e. $1+y \ge 0, \; 2(1+y)\ge z, \; 1+x+y\ge z, $. 
Clearly the first inequality holds.  $z^2 \le 16y\le 4(1+y)^2$ gives you the second inequality, and $z^2\le 4(1+x)y \le (1+x+y)^2$ gives you the last.

$^\dagger$ I have forgotten a good reference for this theorem, except that it can be proved using mixing variables or $uvw$ method.  Will add the reference if I can find it.
