Short version of the question
Consider \begin{equation} g(X_n)=\frac{\mu^2(X_N)}{\mu^2(X_N)+\sigma^2(X_N)}\\ X_N=\{x_1,x_2,...,x_n\},n>1,\forall i:x_i\in \mathbb{N}^+,\exists(i,j):x_i\neq x_j \end{equation} where $\mu,\sigma$ are mean and standard deviation of a random sample $X_n$, respectively.
I would like to find the smallest $U$ so that $g(X_n)<U<1$.
Detailed version of the question
For finding the complexity of a classifier that I have recently proposed, I should find the root of the following function, regarding $z$ \begin{align*} &f(z)-g(X_n),\\ &g(X_n)=\frac{\mu^2(X_N)}{\mu^2(X_N)+\sigma^2(X_N)}\\ &z\in(0,\infty),f(z)\in(0,1),\lim_{z->0^+}{f(z)}=0,\frac{d}{dz}f(z)>0\\ &X_N=\{x_1,x_2,...,x_n\},\forall i:x_i\in \mathbb{N}^+,\exists(i,j):x_i\neq x_j \end{align*} As it can be seen, due to the properties of $X_n$, we have $0<g(X_n)<1$.
However, if $g(X_n)$ has an upper bound like $U$ (i.e. $g(X_n)<U<1$), then I can find the computational complexity of solving $f(z)$ with no problem. But if $g(X_n)$ is free to tend $1^-$, finding the computational complexity becomes problematic.
However, it seems to me that because of the properties of $X_n$ such as having only positive integers and necessarily including at least two distinct members, the upper bound $U$ exists for $g(X_n)$.
Can anyone, please, provide a hint for finding such upper bound?
Such $U$ does not exist. For any given $y_0$ there exist an $X_N$ where $\mu(X_n)\geq1$ and $\sigma^2(X_n)<y_0$.