I want to minimize the function
$$ \frac{x}{1-x^2} + \frac{y}{1-y^2} + \frac{z}{1-z^2} $$ subject to the constraint $$x^2 + y^2 + z^2 = 1 \space\text{and} \space x,y,z > 0$$
Wolfram Alpha tells me that the minimum occurs at $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{3\sqrt{3}}{2})$. I tried to solve this without using Lagrange Multipliers by using AM-GM and Cauchy-Schwarz, but I couldn't figure out how to do it and am interested in a solution.
We can write the inequality to prove as $$\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}\ge \frac{3\sqrt3}2 \tag{$\star$}$$ As equality is achieved for $x=y=z=\frac1{\sqrt3}$, if $(\star)$ holds then we have established the minimum.
Consider function $f(t) = \dfrac{t}{1-t^2}-\frac{\sqrt 3}2 -\frac{\sqrt 3}2(3t^2-1)$. To show the inequality $(\star)$ holds, it is enough to show $f(t)\ge 0$, as the inequality is equivalent to $f(x)+f(y)+f(z) \ge 0$.
Now $f(t) = \dfrac{t(3\sqrt3 t^3-3\sqrt3t+2)}{2(1-t)(1+t)}$, so it is enough to show that $3\sqrt3 t^3+2 \ge 3\sqrt3t$ for $t \in (0, 1)$. But this follows from AM-GM as $3\sqrt3t^3+1+1 \ge 3\left(3\sqrt3 t^3 \times 1 \times 1 \right)^{1/3}=3\sqrt3t$.
Hint: Try with polar coordinates: $$\begin{align} x &= \sin a \sin b\\ y &= \sin a \cos b\\ z &= \cos a\\ 0&\le a,b \le \frac\pi 2 \end{align}$$
