How prove this inequality $6+5|x+y+z|+2|yz+xz+xy|\ge 3(|x|+|y|+|z|)$ let  $x,y,z$ be real numbers,show that

$$6+5|x+y+z|+2|yz+xz+xy|\ge 3(|x|+|y|+|z|)\cdots\cdots\cdots (1)$$

I think this inequality is nice,and I have see this 

$$1+|x+y+z|+|xy+yz+xz|+|xyz|\ge\dfrac{\sqrt[3]{2}}{2}(|x|+|y|+|z|)\cdots\cdots(2)$$
  for any complex numbers $x,y,z$

(2) This solution:
let
$1\ge|x|\ge |y|\ge |z|,y=bx,z=cx,1\ge|b|\ge |c|$,and let $\theta=|x|\le 1$
then

$$LHS=1+\theta|1+b+c|+\theta^2|b+c+bc|+\theta^3|bc|-\dfrac{\sqrt[3]{2}}{2}\theta(1+|b|+|c|)$$
  then
  $$LHS\ge 1+\theta^3(|1+b+c|+|b+c+bc|+|bc|)-\dfrac{\sqrt[3]{2}}{2}\theta(1+|b|+|c|)\ge1+\theta^3-\dfrac{3\sqrt[3]{2}}{2}\theta\ge 0$$

when $|x|,|y|,|z|\ge 1$ is obvious.Done!
But this $(1)$ I can't prove it.Thank you 
 A: By switching all the signs, we can clearly assume that at least 2 variables are non-negative. 
If $x,y,z\geq 0$, there is nothing to prove, so we only need to prove that:
$$\forall x,y,z\geq 0,\quad LHS=6+5|x+y-z|+2|xy-xz-yz|\geq 3(x+y+z)=RHS. $$
Now there are 3 cases. If $z\leq\frac{xy}{x+y}=\frac{1}{2}HM(x,y)$, then:
$$ LHS-RHS \geq 6+5(x+y-z)-3(x+y+z)\geq\frac{2}{x+y}\left((x+y)^2-4xy\right)\geq 0.$$
If $z\geq x+y = 2 AM(x,y)$, then:
$$ LHS-RHS \geq 6+2x+2y+2z(x+y-4)-2xy \geq \frac{3}{2}(x+y-2)^2 \geq 0.$$
If $\frac{1}{2}HM(x,y)=\frac{xy}{x+y}\leq z \leq x+y = 2 AM(x,y)$, then:
$$ LHS-RHS = 2\left((x+y-4)z+(x+y+3-xy)\right), $$
so we have to prove that $\frac{xy-x-y-3}{x+y-4}$ cannot belong to the interval $\left(\frac{xy}{x+y},x+y\right)$ for any couple $(x,y)$ of non-negative real numbers. If $x+y\geq 4$, proving $\frac{xy-x-y-3}{x+y-4}<\frac{xy}{x+y}$ is equivalent to prove $-3(x+y)-(x+y)^2+4xy<0$, that is trivial since the LHS is $\leq -12$. If $x+y<4$, proving $\frac{xy-x-y-3}{x+y-4}>x+y$ is equivalent to prove $xy+3(x+y)-3<(x+y)^2$, that is a consequence of $(x+y-2)^2>0$ since $xy\leq\frac{1}{4}(x+y)^2$. The proof is complete.
