Pinko wrote: Prove that if $x$, $y$, and $z$ are non-negative numbers and $x^2+y^2+z^2=1$, then the following inequality is true: $$\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2 }\geq \frac{3\sqrt{3}}{2}$$ See also here https://artofproblemsolving.com/community/c6h600016p3561899

The function $f(x)=\frac{1}{1-x^2}$ is convex for $0<x<1$, which is obviously the case. Then we use Jensen inequality: $\frac{x}{x+y+z} \frac{1}{1-x^2}+\frac{y}{x+y+z} \frac{1}{1-y^2}+\frac{z}{x+y+z} \frac{1}{1-z^2} \geq f( \frac{x^2+y^2+z^2}{x+y+z})= $ $= f( \frac{1}{x+y+z}) = \frac{1}{1-\frac{1}{(x+y+z)^2}} = \frac{(x+y+z)^2}{(x+y+z)^2-1}$ Now denote $x+y+z=a$. We have $a^2>x^2+y^2+z^2=1$ so $a>1$ and define $g(x) = \frac{x^3}{x^2-1}$ $LHS \geq \frac{a^3}{a^2-1} = g(a)$ By taking first derivative of $g(a)$ we get the function is decreasing on $(1,\sqrt{3})$ and increasing on $(\sqrt{3},\infty)$ so we conlude $LHS \geq g(a) \geq g(\sqrt{3}) = \frac{3\sqrt{3}}{2}$