1, 1, s),$ OPT$(1,0,0, 1,0,0, s)=1\cdot (1/1)^Q \cdot 3=3=$OPT$_3(1, 0, 0,s).$ The following lemma shows the idea of OPT.

\[LEMOPT\] Given $\bar{\lambda}>0$ and let $\lambda$ be the maximum of the $\lambda_i.$ Let $ \,  a_i, c_i \,>0$ be such that $P_i:= a_ic_i= \lambda_i/\lambda.$ Then $$\begin{aligned}
\Psi:= \Psi(\bar{\omega}, \bar{\lambda}, \bar{a}\cdot s, \bar{c}) \le \frac{1}{3^\gamma} 
\mbox{OPT}(\bar{a}, \bar{c}, s).       \nonumber \end{aligned}$$

The factors of $\Psi$ one by one: The first factor: The AGM-inequality gives\
$ \prod_{i=0, 1, 2} \left(\frac{q(a_i s)}{\omega_i q(s)} \right)^{\omega_i} \,\le \, \mbox{OPT}_1(\bar{a}, s). $ (Applies for $\omega_i=0$, too.)

The second factor: We have $P_0+P_1+P_2= a_0c_0+ a_1c_1+ a_2c_2 = 1/\lambda$ and $\lambda_i/a_ic_i= \lambda$ for $i=0, 1, 2.$ Recall $Q=k\gamma, $ and the second factor of $\Psi \,=$ $$\begin{aligned}
\prod_{i=0, 1, 2} \left( \frac{\lambda_is}{a_isc_i}\right)^{\lambda_ik\gamma} \,= \, \lambda^{k \gamma} = 
\left(\frac{1}{a_0c_0+a_1c_1+a_2c_2}\right)^Q\,= \mbox{OPT}_2(\bar{a}, \bar{c}, s). \nonumber \end{aligned}$$

The third factor: We let $C_1=\sum_i c_i$ and $C_2= \sum_i c_i^2 -c_0c_1-c_0c_2-c_1c_2.$ Then $r(\bar{c}) =|r(\bar{c})| \le (1/3)(C_1^k\,+ \,2 C_2^{k/2})$ by the triangle inequality and as $|c_0 +{\bf w_1}c_1 + {\bf w_2}c_2|= [(c_0-1/2\cdot (c_1+c_2))^2+ (\sqrt{3}/2( c_1-c_2))^2]^{1/2}=C_2^{1/2}.$ Then $|r(\bar{c})|^\gamma  \le 1/3^\gamma (C_1^k+2C_2^{k/2})^\gamma \le 1/3^\gamma (C_1^{k\gamma}+2^\gamma C_2^{\gamma k/2})\le1/3^\gamma \mbox{OPT}_3(\bar{c}, s)$ as $Q=k\gamma,$ and as $x^\gamma$ is concave (by $\gamma <1)$ we have $(y+z)^\gamma \le y^\gamma + z^\gamma.$

The following picture 