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For this equal-strength power-law mutual repression, the logarithmic gains and their product at equilibrium areThe mutual repression stability criterion immediately excludes multistability if $mn\le 1$, since $P<mn$ at every finite positive equilibrium. The common prefactor $\lambda$ gives a stronger exclusion: if either exponent is at most one, multistability is impossible for every $\lambda$, including some cases with $mn>1$.

To prove the additional cases, suppose $m\le 1<n$. At equilibrium the two expressions for $\lambda$ implyPut $a=x/(1+x)$ and $b=y/(1+y)$. Because $m\le 1$, the left side is at least $a$, so $a\le b(1-b{)}^{n-1}$. It follows thatThe maximum in the second inequality occurs at $b=2/(n+1)$. Its strict bound follows from $4n/(n+1{)}^2\le 1$, $m\le 1$, and $((n-1)/(n+1){)}^{n-1}<1$. Interchanging the components proves the case $n\le 1<m$; the case with both exponents at most one was already covered by $mn\le 1$. Every root of $F(x)=f(g(x))-x$ therefore has $F^{\prime }<0$. Since $F$ begins positive and ends negative, there is exactly one root: two downward crossings would require another crossing with nonnegative derivative between them. The unique equilibrium is stable.

For completeness, neither exponent can be relaxed above one in this exclusion. If $m>1$ and $n>1$, take $\lambda$ sufficiently large. On $x\in [\lambda /2,\lambda ]$, $g(x)=O({\lambda }^{1-m})\to 0$, so $f(g(x))=\lambda (1+o(1))$ maps that interval into itself. It has a fixed point withIt is a stable equilibrium. By exchanging the components, there is a second stable fixed point with $y\sim \lambda$ and $x\sim {\lambda }^{1-n}$. These two regions are disjoint for large $\lambda$, proving bistability for some production strength. Hence the full answer isFor example, $m=1/2,n=3$ has $mn>1$ but still a unique stable equilibrium for all $\lambda$. Thus the product-only necessary condition for strong feedback is not sufficient when both production prefactors are constrained to be equal.