= Solution
For
$$
M=
\begin{pmatrix}
-1&-1\\
1+\delta&1-\delta
\end{pmatrix},
$$
we have
$$
\operatorname{tr}M=-\delta<0,
\qquad
J=\det M=2\delta>0,
$$
so the homogeneous equilibrium is stable for $\delta>0$. The Turing condition is
$$
-1+d(1-\delta)>2\sqrt{2\delta d}.
$$
Writing $y=\sqrt d$, equality at threshold is
$$
(1-\delta)y^2-2\sqrt{2\delta}\,y-1=0.
$$
Its positive root is
$$
y_c
=\frac{\sqrt{2\delta}+\sqrt{1+\delta}}{1-\delta},
\qquad
\boxed{
d_c
=\left(
\frac{\sqrt{2\delta}+\sqrt{1+\delta}}{1-\delta}
\right)^2
}.
$$
For $d>d_c$ sufficiently close to the threshold the strict instability condition holds, while
$$
d_c=1+2\sqrt{2\delta}+O(\delta)\longrightarrow1
\qquad(\delta\to0^+).
$$
Thus unequal diffusivities are necessary, but their ratio can be arbitrarily close to one when the stable reaction matrix is itself arbitrarily close to marginal stability. This is the <near-unity diffusivity ratio for a Turing instability>.
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