exist $y_1 \ge  y_2$ with $P_i= A(y_i)\cdot y_i.$ We represent $y_i$ such that Lemma \[lem1\] is applicable. $$\begin{aligned}
y\,:= \, \frac{y_1 + y_2}{2} ,\,\,  z \,:= \, \frac{y_1 - y_2}{2}.
\mbox{ Then } y_1 = y+z , y_2 = y-z , 0 \le z \le y. \nonumber \end{aligned}$$ We show $y\le \frac{1}{2Q}$ and Lemma \[lem1\] applies to $y, z.$ $$\begin{aligned}
\frac{7}{20Q}\, \ge \, P_1+P_2 \,= \, \frac{7}{10}Q (y_1^2+ y_2^2) \Longrightarrow y_1^2+ y_2^2 \,\le \, \frac{1}{2Q^2}. \nonumber \\
 (y_1+ y_2)^2 \le  2 y_1^2+2 y_2^2 \le  \frac{1}{Q^2} \mbox{ and } y= \frac{y_1+y_2}{2}\le \frac{1}{2Q}. \nonumber \end{aligned}$$ With $a_0=c_0=1, a_1=A(y+z), a_2=A(y-z), c_1=y+z, c_2=y-z$ we have $a_ic_i=P_i.$ By Lemma \[LEMOPT\] $\Psi:= \Psi(\bar{\omega},\bar{\lambda}, \bar{a}\cdot s, \bar{c})\le 1/3^\gamma \mbox{OPT}, \,\,
\mbox{OPT}:= \mbox{OPT}(\bar{a}, \bar{c},s).$ If $P_1 \ge \varepsilon $ for an $\varepsilon>0$ we have OPT$< 3-\delta'$ by Lemma \[lem1\] and Theorem \[OPT\] holds.

For smaller $P_1$ we have OPT $ \le 3,$ approaching $3$. Only (1) of Theorem \[OPT\] holds. To get (2) for small $P_1$ we argue as follows: For $P_1$ approaching $0$ we see that $c_1$ and $c_2$ approach $0 .$ We consider the treatment of the factor $r(\bar{c})$ in the proof Lemma \[LEMOPT\]. Both $C_1$ and $ C_2$ from this proof approach $1$ in this 