First, we know that the $3$ number $-1, \frac12, 2$ is a solution, so $k\geq\frac32$. Assume that there is a solution $>\frac32$. Suppose the $3$ numbers are $a, b, c$, where $c<a<b$. There can't be $3$ positive numbers, otherwise either $c, a\in(0, 1]$ or $\frac1a, \frac1b\in(0, 1]$, and the solution won't be $>\frac32$. Similarly, there can't be $3$ negative numbers. Since $a, b, c$ is a solution $\iff -a, -b, -c$ is a solution $\iff\frac1a, \frac1b, \frac1c$ is a solution. $\therefore$ WLOG suppose that $c<0<a<b$ and $ab\leq1$. If $a=1$, then $\frac1a-\frac1b<\frac32$; if $b=1$, then $b-a<\frac32$. $\therefore c<0<a<1<b$. $b\geq a+k$ $c\leq a-k$ $\frac1c\leq\frac1b-k\leq\frac1{a+k}-k$ Since $c<0$, and $a<1<k$, $\frac1{a+k}<1<k$, we have $(a-k)(\frac1{a+k}-k)\leq c\times\frac1c=1$. $\Rightarrow k^3+ak^2-k-ak^2-a^2k+a\leq a+k$. $\Rightarrow k^3-a^2k\leq2k$ $\Rightarrow k^2-a^2\leq2$ $\therefore a\geq\sqrt{k^2-2}$ $\Rightarrow b\geq\sqrt{k^2-2}+k$ From the assumption, $1\geq ab\geq\sqrt{k^2-2}(\sqrt{k^2-2}+k)=k^2-2+k\sqrt{k^2-2}$. $\Rightarrow3-k^2\geq k\sqrt{k^2-2}$ $\Rightarrow3-k^2\geq0$, and therefore $k^4-6k^2+9\geq k^4-2k^2$. $\Rightarrow4k^2\leq9$ $\therefore k\leq\frac32$, and we finish the proof.