= Solution
Put $q=u\cdot v$ and $d=u+v$. The <linear independence> assumption implies $d\ne0$. Since $w$ is a <unit vector>, the <Cauchy-Schwarz inequality> gives
$$
S=q+w\cdot d\ge q-\|d\|.
$$
Equality holds exactly when $w$ points opposite to $d$. Its required unit length then fixes the scale:
$$
\boxed{w=-\frac{u+v}{\|u+v\|},\qquad\lambda=-\frac1{\|u+v\|}=-\frac1{\sqrt{2+2q}}.}
$$
Here $\|u+v\|^2=2+2q$ because both vectors have unit norm. The minimizing vector is unique, and the <minimum pairwise dot-product sum of three unit vectors> for these fixed $u,v$ is
$$
\boxed{S_{\min}=q-\sqrt{2+2q}.}
$$
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