= Solution
The <Cauchy point> minimizes the model along the negative <gradient> ray inside the <trust region>. With $q=g^THg$, write
$$
s_C=-\tau g,\qquad m(-tg)=a-t\|g\|^2+\tfrac12t^2q,\qquad 0\leq t\leq\Delta/\|g\|.
$$
If $q\leq0$, this last <quadratic function> decreases throughout the interval. If $q>0$, its unconstrained <minimum> occurs at $t=\|g\|^2/q$. Thus
$$
\boxed{\tau=\begin{cases}\Delta/\|g\|,&g^THg\leq0,\\\min\{\Delta/\|g\|,\|g\|^2/(g^THg)\},&g^THg>0.\end{cases}}
$$
The additive constant $a$ has no effect on the <Cauchy point>. To obtain the <Cauchy decrease bound>, first suppose $q>0$. Since $\tau q\leq\|g\|^2$,
$$
a-m(s_C)=\tau\|g\|^2-\tfrac12\tau^2q\geq\tfrac12\tau\|g\|^2.
$$
The <spectral norm> inequality $q\leq\|H\|\|g\|^2$ gives $\tau\geq\min\{\Delta/\|g\|,1/\|H\|\}$. Hence
$$
\boxed{a-m(s_C)\geq\tfrac12\|g\|\min\{\Delta,\|g\|/\|H\|\}}.
$$
If $q\leq0$, the decrease is at least $\Delta\|g\|$, which also implies the displayed bound. When $H=0$, interpret $\|g\|/\|H\|=+\infty$. Thus the universal choice is $c=1/2$.
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