such that $$\begin{aligned}
&\left|\frac{4 f f_{ss}}{g^2}\right| + \left|\frac{4B f_s^2 f_{ss}}{f}\right| + \left|\frac{8B f_s g_s f_{ss}}{g}\right|
+ \left|\frac{2g f_s g_s f_{ss}}{f^2}\right|\\
&\quad\leq\left(\frac 12 f_{ss}^2 + \frac{C' f^2}{g^4}\right) + \left(\frac{B}{2}\, f_{ss}^2 + C' B \frac{f_s^2}{f^2}\right)
+ \left(\frac B2 f_{ss}^2 + C'B \frac{g_s^2}{g^2}\right) + \left(\frac 12 f_{ss}^2 + C'\frac{f_s^2}{f^2}\right) \\
&\quad\leq(B+1) f_{ss}^2 + C'(B+1)\left( \frac{f_s^2}{f^2} + \frac{g_s^2}{g^2} +\frac{f^2}{g^4}\right).\end{aligned}$$ The remaining terms in  can be estimated in a similar manner. Thus we find that $$\begin{split}
\frac{d}{dt} Q_{\min} &\ge  2Bf_{ss}^2 +  \frac{\bar{C} f^2}{2 g^4} + \frac{\bar{C} f_s^2}{2f^2} + \frac{\bar{C} A g_s^2}{g^2} \\
&\qquad- C'(1 + A + B) \left(\frac{f^2}{g^4} + \frac{f_s^2}{f^2} + \frac{g_s^2}{g^2}\right) - (B + 1) f_{ss^2} \\
&\ge 0,
\end{split}$$ if we choose $A = 1$, $B = 2$ and $\bar{C}$ sufficiently large so that $\bar{C} > C'(1 + A + B)$. Therefore, in this case, either $Q \ge -\bar{C}$ or $\frac{d}{dt}Q_{\min} \ge 0$.

\[case-2\] A minimum of $Q$ occurs on ${\mathcal}S^2_\pm$.

The only difference from Case \[case-1\] is that one must deal with the term $\frac{Q_s f_s}{f}$ at ${\mathcal}S^2_\pm$. We apply l’Hôpital’s rule to see that $$\frac{Q_s f_s}{f}\Big|_{{\mathcal}S^2_{\pm}}
= \left(Q_{ss} + Q_s \frac{f_{ss}}{f_s}\right)\Big|_{{\mathcal}S^2_{\pm}}
 = Q_{ss}\big|_{{\mathcal}S^2_{\pm}}.$$ However, smoothness of either function $Q_{\pm}(s,\cdot) :=Q(s - s_{\pm}, \cdot)$ at a minimum on ${\mathcal}S^2_\pm$ shows that 