\, \,$, and $  \frac{7}{20} \le A \le 1-\frac{1}{Q}.$ Then\
OPT$(z):= $ OPT$\left(1, A,\, A, \, 1, 1/(2Q)+z, \, 1/(2Q)-z \,  , \, s \right) \le 3 \,- \, \delta$ for $0 \le z \le 1/(2Q).$

\[lem3\] Let $\, s\, \ge \, 7,$ and $  1/(2Q) \le C \le 1/2.$ Then\
OPT$(z):= $ OPT$\left(1, 1-1/Q,\, 1-1/Q, \,1,  C+z, \, C-z\,, s \right)\, \le  3 \,- \, \delta$ for $0 \le z \le C.$

\[lem4\] Let $s\ge 15$ and $  A(x)\,=\, A(x, s) \, := \, 1+ 7/(10Q)\cdot x \,- \, 7/(10Q).$\
(a) OPT$(y):= $OPT$\left(1, A(y), A(y), 1, y, y, s \right) $ is strictly increasing in $4/10 \le y \le 1.$ The final value is OPT$(1)=3.$\
(b) Given $4/10 \le y \le 1,$ OPT$(z):=$OPT$\left(1, A(y+z), A(y-z),1,  y+z \, , y-z \, , \, s \right)   $ is decreasing in $0 \le z \le \min\{y, 1-y\}.$

[*Proof of Theorem \[OPT\] from the preceding lemmas.*]{} We prove Theorem \[OPT\] for $\lambda_0 \ge \lambda_1 \ge \lambda_2>0 $ first. We denote $P_i:=\lambda_i/ \lambda_0, \, \mbox{ then }  1\ge P_1\ge P_2>0.$

[*Case 1: $P_1\, +\, P_2 \le \frac{7}{20 Q}$.*]{} With $A(x)$ from Lemma \[lem1\] we have $A(x)\cdot x \,= \, (7/10)Q \cdot x^2.$ Thus there 