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 $\text{min}(\dfrac{x}{z})=?$

Question

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?

Answer 1

Since $x \ge y \ge z > 0$, we have $x \ge x - y + z > 0$. This leads to

$$\begin{align}x^2 \ge (x-y+z)^2 &= x^2+y^2+z^2 -2(xy+yz-xz)\\ &= 4xz + (x^2+y^2+z^2 - 2(xy+yz+xz)) \\ &\ge 4xz\end{align}$$ As a result, $\displaystyle\;\frac{x}{z} \ge 4$. Notice this ratio $4$ is achievable at $(x, y, z) = (4, 1, 1)$ which satisfies the constraint $x^2+y^2+z^2 \ge 2(xy+yz+zx)$, this is the minimal value of $\displaystyle\;\frac{x}{z}$.

Update

About the side question how I have solved this problem. I didn't solve it by algebra. The first thing I do is figure out what are the domain of $(x,y,z)$ subject to the constraints:

1. $x \ge y \ge z > 0$.
2. $x^2+y^2+z^2 \ge 2(xy+yz+xz) \iff 2(x^2+y^2+z^2) \ge (x+y+z)^2$.

Notice the constraints are homogeneous, i.e. if $(x,y,z)$ satisfy the constraints, so do any scalar multiple $(\lambda x,\lambda y,\lambda z)$ for $\lambda > 0$. This means we can restrict out attention to those $(x,y,z)$ lies on the plane $x + y + z = 1$.

Constraint $1$ tell us the $(x,y,z)$ belongs to one-sixth of the equilateral triangle with vertices $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. Constraint $2$ tell us $(x,y,z)$ is lying outside a circle on that equilateral triangle. The resulting domain is a triangular like shape bounded by two straight lines and a circular arc.

If one play around the values of $\frac{z}{z}$ along the boundary of this domain, one will find the minimum is achieved at the vertex $(\frac23,\frac16,\frac16)$ with value $4$. Knowing the answer is $4$, one can look back at the algebra and try to find a way to get that. The rest is really trial and error.

In short, in order to solve this problem, I use following sort of well known approach:

First guess the right answer by whatever means and then justify it.

Answer 2

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.