Given $s,i \in [0,1]$ I need to check if this system can be verified:
$$\begin{cases} s = 1 - \frac{\min\{x,y,z\}}{i} \\ \frac{x+y+z}{3} = i \end{cases}$$
where $x,y,z \in [0,1]$ are unknown.
For example if $s=1$ and $i=1$ the system is not verified since $i$ should be the average of three real numbers in $[0,1]$, hence $\min\{x,y,z\} = 1$ and $s$ should be $1-1 = 0 \neq 1$.
The problem is symmetric in $\,x,y,z\,$ so it can be assumed WLOG that $\,x \le y \le z\,$, then $\,\min\{x,y,z\} = x\,$ and the system becomes:
$$ \begin{cases} s = 1 - \frac{x}{i} \\ \frac{x+y+z}{3} = i \end{cases} \;\;\iff\;\; \begin{cases} si = i - x \\ x+y+z = 3i \end{cases} \;\;\iff\;\; \begin{cases} \begin{align} x = (1-s)i \quad\tag{1} \\ y+z = (s+2)i \quad\tag{2} \end{align} \end{cases} $$
The first equation $(1)$ is always satisfiable since the RHS $\,(1-s)i \in [0,1]\,$ is within the allowed range for $\,x\,$.
For $\,(2)\,$ to have eligible solutions $\,y,z \in [x, 1]\,$, the necessary and sufficient condition is that $\,y+z \in [2x, 2] \iff 2x \le y+z \le 2\,$, which by $\,(1)$-$(2)\,$ is:
$$\require{cancel} 2(1-s)\,i \;\le\; (s+2)\,i \;\le\; 2 \;\;\iff\;\; \begin{cases} \begin{align} 2(1-s)\le s+2 \;\;&\iff\;\; 0 \le 3s \quad\tag{3} \\[5px] (s+2)i \le 2 \;\;& \tag{4} \end{align} \end{cases} $$
Condition $(3)$ is always satisfied, so in the end the condition for solvability is $(4)\,$.