# Minimum of $\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}+\frac{z}{\sqrt{1-z^2}}$

Let $0\leq x,y,z<1$ and $x^2+y^2+z^2=1$. What is the minimum value of $$\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}+\frac{z}{\sqrt{1-z^2}}?$$

From the condition, the point $(x,y,z)$ lies on a sphere with radius $1$. At the equality point, all three variables are equal to $1/\sqrt{3}$, and the value taken is $3/\sqrt{2}$. If one variable is $0$ and the other two are equal, the two variables are equal to $1/\sqrt{2}$, and the value taken is $2$.

• You have the wrong value when the three variables are equal, it should be $\frac{3}{\sqrt{2}}$, which is bigger than $2$. Apr 1, 2018 at 9:04

By AM-GM $$\sum_{cyc}\frac{x}{\sqrt{1-x^2}}=\sum_{cyc}\frac{2x^2}{2\sqrt{x^2(1-x^2)}}\geq\sum_{cyc}\frac{2x^2}{x^2+1-x^2}=2.$$ The equality occurs for $z=0$ and $x=y=\frac{1}{\sqrt2},$ which says that $2$ is a minimal value.

You may convert it to become a function of two variables, the find the minimum value of that function. To do this, use

$$z = \sqrt{1-(x^{2}+y^{2})}$$

And substitute this in

$$\frac{x}{\sqrt{1-x^{2}}} + \frac{y}{\sqrt{1-y^{2}}} + \frac{z}{\sqrt{1-z^{2}}}$$

so the function will be

$$f(x,y) = \frac{x}{\sqrt{1-x^{2}}} + \frac{y}{\sqrt{1-y^{2}}} + \frac{ \sqrt{1-(x^{2}+y^{2})} }{ \sqrt{x^{2}+y^{2}} }$$

Use partial derivatives to find the minimum value..?