= Solution
For a spatially uniform <steady state> with nonzero <bacteria> abundance, $b(1-b)=0$ forces $b_*=1$. The <nutrient> balance then gives
$$
\boxed{(b_*,n_*)=\left(1,\frac{\lambda}{\mu(1+b_0^2)}\right).}
$$
To test spatially uniform <linear stability>, write
$$
F(b,n)=b(1-b),\qquad G(b,n)=\frac{\lambda b^2}{b_0^2+b^2}-\mu bn.
$$
At the positive <equilibrium>, the <Jacobian matrix> is
$$
J_0=\begin{pmatrix}-1&0\\\eta&-\mu\end{pmatrix},\qquad
\eta=\frac{2\lambda b_0^2}{(1+b_0^2)^2}-\mu n_*
=\frac{\lambda(b_0^2-1)}{(1+b_0^2)^2}.
$$
Its <eigenvalues> are $\boxed{-1,\ -\mu}$, both negative, so the positive <equilibrium> is asymptotically stable against spatially uniform perturbations. This remains true at $\mu=1$, where a repeated <eigenvalue> may produce a decaying $te^{-t}$ contribution. The printed <stability> claim must be understood in this spatially uniform sense: sufficiently strong <chemotaxis> can destabilize nonuniform perturbations, as the next answer shows.
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