show that $x+y+z\ge 2$ 
Let $x,y,z\ge 0$. If
  $$x^{2016}y^{2016}+y^{2016}z^{2016}+x^{2016}z^{2016}=1$$ then show
  that $$x+y+z\ge 2$$

I have tried
$$(x+y+z)^2-3(xy+yz+xz)=\dfrac{1}{2}\left[ (x-y)^2+(y-z)^2+(z-x)^2\right] \ge 0$$
so we have
$$(x+y+z)^2\ge 3(xy+yz+xz)$$
but I am stuck here. How could I continue this proof?
 A: For $z=0$ and $x=y=1$ we get a value $2$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\left(\frac{x+y+z}{2}\right)^{4032}\geq\sum_{cyc}x^{2016}y^{2016},$$
for which we'll prove the following.
Let $x$, $y$ and $z$ be non-negative numbers and nutural $n\geq2$. Prove that:
$$\left(\frac{x+y+z}{2}\right)^{2n}\geq x^ny^n+x^nz^n+y^nz^n.$$
1. $n=2$.
We need to prove that 
$$(x+y+z)^4\geq16(x^2y^2+x^2z^2+y^2z^2).$$
Indeed, by AM-GM
$$(x+y+z)^4=\left(\sum_{cyc}(x^2+2xy)\right)^2\geq8\sum_{cyc}x^2\sum_{cyc}xy=$$
$$=8\sum_{cyc}(x^3y+x^3z+x^2yz)\geq8\sum_{cyc}(x^3y+x^3z)\geq8\sum_{cyc}(2x^2y^2)=16(x^2y^2+x^2z^2+y^2z^2).$$
2. Now, $$\left(\frac{x+y+z}{2}\right)^{2k+2}=\left(\frac{x+y+z}{2}\right)^4\left(\frac{x+y+z}{2}\right)^{2k-2}\geq$$
$$\geq\sum_{cyc}x^2y^2\sum_{cyc}x^{k-1}y^{k-1}=\sum_{cyc}(x^{k+1}y^{k+1}+x^{k+1}y^{k-1}z^2+x^{k+1}z^{k-1}y^2)\geq$$
$$\geq x^{k+1}y^{k+1}+x^{k+1}z^{k+1}+y^{k+1}z^{k+1}.$$
Thus, by induction for ending of the proof it remains to prove that
$$(x+y+z)^6\geq64(x^3y^3+x^3z^3+y^3z^3).$$
We can make it by the similar way:
$$(x+y+z)^6=(x+y+z)^4\sum_{cyc}(x^2+2xy)\geq16\sum_{cyc}x^2y^2\sum_{cyc}(x^2+2xy)=$$
$$=16\sum_{cyc}(x^4y^2+x^4z^2+x^2y^2z^2+2x^3y^3+2z^3y^2z+2x^3z^2y)\geq$$
$$\geq16\sum_{cyc}(x^4y^2+x^4z^2+2x^3y^3)\geq16\sum_{cyc}(2x^3y^3+2x^3y^3)=64\sum_{cyc}x^3y^3$$
and we are done!
