Solve for $x,y,z \in \mathbb{R^+}$ ,$x^2+y^2+z^2=xyz+4$ and $xy+yz+zx=2(x+y+z)$ 
Find all positive real numbers such that
  $$x^2+y^2+z^2=xyz+4$$
  And $$xy+yz+zx=2(x+y+z)$$.

I substitute $x,y,z$ by $(a+\frac{1}{a}),(b+\frac{1}{b}),(c+\frac{1}{c})$ respectively where $abc=1$ which comes from the first relation. Putting the values in the second relation it becomes complex.
I tried another way. Multiplying second Equation by 2 and adding with the first equation we get ,$(x+y+z-2)^2=xyz+8$.I failed to find any method. Any ideas guys?

After homogenization we obtain:
$$\frac{(x^2+y^2+z^2)(xy+xz+yz)}{2(x+y+z)}=xyz+\frac{(xy+xz+yz)^3}{2(x+y+z)^3}.$$
We'll prove that
$$\frac{(x^2+y^2+z^2)(xy+xz+yz)}{2(x+y+z)}\geq xyz+\frac{(xy+xz+yz)^3}{2(x+y+z)^3}.$$
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$, and $xyz=w^3$.
We see that it's enough to prove this inequality for a maximal value of $w^3$, which happens for equality case of two variables.
Since the inequality is symmetric and homogeneous, we can assume $y=z=1$, which gives
$$(2x+7)(x^2-1)^2\geq0.$$
The equality occurs for $x=y=z,$ which gives
$$3x^2=6x$$ or $$x=y=z=2.$$
Another way.
We need to prove that: $$\frac{(9u^2-6v^2)\cdot3v^2}{6u}\geq w^3+\frac{27v^6}{54u^3}.$$
Now, by AM-GM $$v^4\geq uw^3,$$ which says that it's enough to prove that
$$\frac{(9u^2-6v^2)v^2}{2u}\geq\frac{v^4}{u}+\frac{v^6}{2u^3}$$ or
$$9u^4-8u^2v^2-v^4\geq0$$ or
$$(u^2-v^2)(9u^2+v^2)\geq0$$ and since $$u^2-v^2\geq0$$ it's $$\sum_{cyc}(x-y)^2\geq0,$$  our inequality is true.
