show that $x+y+z\ge 2$

Question

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?

Answer 1

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!