= Solution
Set the time derivatives to zero. The scalar equations give
$$
b=\frac{a}{1+a^2},\qquad c=\frac{a^2}{1+a^2},\qquad d=\frac{a\tau}{\tau^2+a^2},\qquad e=\frac{a^2}{\tau^2+a^2}.
$$
For the nonzero convection branch, division of the velocity equation by $a$ gives
$$
\boxed{r(x)=(1+x)+\frac{r_s\tau(1+x)}{\tau^2+x},\qquad x=a^2\ge0.}
$$
Its zero-amplitude limit is $r_e=1+r_s/\tau$. Differentiation yields
$$
r'(x)=1-\frac{r_s\tau(1-\tau^2)}{(\tau^2+x)^2}.
$$
For $0<\tau<1$, the sharp condition for a <subcritical bifurcation> at the steady onset is $r_s>\tau^3/(1-\tau^2)$. The stronger bound printed in the question is sufficient, since $(1-\tau)^2<1-\tau^2$. Under it the branch initially runs toward smaller $r$, and the unique <saddle-node bifurcation> has
$$
\boxed{x_*=a_*^2=-\tau^2+\sqrt{r_s\tau(1-\tau^2)}.}
$$
It is a genuine positive-amplitude minimum: $r''(x)=2r_s\tau(1-\tau^2)/(\tau^2+x)^3>0$. Put $h=\sqrt{r_s\tau}$ and $j=\sqrt{1-\tau^2}$. At the minimum, $1+x_*=j(j+h)$ and $\tau^2+x_*=hj$, so
$$
\boxed{r_{\rm min}=(h+j)^2=\left(\sqrt{r_s\tau}+\sqrt{1-\tau^2}\right)^2.}
$$
These results describe the <steady fold of a thermosolutal Lorenz model>. They do not by themselves assert stability against every oscillatory disturbance.
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