= Solution
Let
$$
M=\begin{pmatrix}f_u&f_v\\g_u&g_v\end{pmatrix}
$$
be the reaction Jacobian at the homogeneous equilibrium. Stability without diffusion gives
$$
f_u+g_v<0,
\qquad
J=\det M=f_ug_v-f_vg_u>0.
$$
For a spatial Fourier mode with $-\nabla^2$ eigenvalue $k^2$, the linearized matrix is
$$
M-k^2\begin{pmatrix}D_u&0\\0&D_v\end{pmatrix}.
$$
Its trace is even more negative than $\operatorname{tr}M$, so instability occurs exactly when its determinant becomes negative. Put $z=D_vk^2$ and $d=D_u/D_v$. Then
$$
\Delta(z)
=J-(f_u+dg_v)z+dz^2.
$$
This upward-opening quadratic is negative for some $z>0$ exactly when its minimum occurs at positive $z$ and lies below zero:
$$
f_u+dg_v>0,
\qquad
J-\frac{(f_u+dg_v)^2}{4d}<0.
$$
The second strict inequality implies the first when written with a positive square root, so the <two-species diffusion-driven instability criterion> becomes
$$
\boxed{f_u+dg_v>2\sqrt{dJ}}.
$$
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