let $x,y ,z \in \mathbb{R^+}$ such that $x \geq y \geq z$ and $x^2+y^2+z^2 \geq 2xy+2xz+2yz$ now find the $\text{min}(\dfrac{x}{z})=?$ 
Let $x,y ,z \in \mathbb{R^+}$ such that $x \geq y \geq z$ and $x^2+y^2+z^2 \geq 2xy+2xz+2yz$. Find $\text{min}\bigg(\dfrac{x}{z}\bigg)= \ ?$

I found $(x-y+z-2\sqrt{xz})(x-y+z+\sqrt{xz})\geq 0$ 
but now what do I do?

Since $$2(xy+xz+yz)-x^2-y^2-z^2)=$$
$$=(\sqrt{x}+\sqrt{y}+\sqrt{z})(\sqrt{x}+\sqrt{y}-\sqrt{z})(\sqrt{x}-\sqrt{y}+\sqrt{z})(\sqrt{y}+\sqrt{z}-\sqrt{x})\leq0,$$
We obtain $$\sqrt{y}+\sqrt{z}-\sqrt{x}\leq0.$$
Thus, $$\frac{x}{z}\geq\frac{(\sqrt{y}+\sqrt{z})^2}{z}\geq4.$$
The equality occurs for $y=z$, which says that we got a minimal value. 
