Question
Let $I$ be a set of indices, and for all $i \in I$, assume $$0 \le {a_i},{b_i},{c_i},{u_i},{v_i} \le 1 \;,$$ with $u_i+v_i=1$.
Assume that $L$ is an upper bound on both ${\sum\nolimits_i {\left| {{a_i} - {c_i}} \right|} }$ and ${\sum\nolimits_i {\left| {{b_i} - {c_i}} \right|} }$.
If $d_i=u_ia_i+v_ib_i$ for all $i \in I$, find an upper bound for ${\sum\nolimits_i {\left| {{d_i} - {c_i}} \right|} }$.
My proof that $2L$ is an upper bound:
Define: $${m_i} = \min \left\{ {{a_i},{b_i}} \right\} \qquad {M_i} = \max \left\{ {{a_i},{b_i}} \right\} \;,$$ and note that ${m_i} \leqslant {d_i} \leqslant {M_i}$. Moreover, let $${\varepsilon _i} = \max \left\{ {\left| {{a_i} - {c_i}} \right|,\left| {{b_i} - {c_i}} \right|} \right\} \;.$$ Then: $$\left. \begin{gathered} \text{If } {d_i} \geqslant {c_i}\,\,\,\, \Rightarrow \,\,\,\,\,\left| {{d_i} - {c_i}} \right| = {d_i} - {c_i} \leqslant {M_i} - {c_i} \leqslant \left| {{M_i} - {c_i}} \right| \leqslant {\varepsilon _i}\\ \text{If } {d_i} < {c_i}\,\,\,\, \Rightarrow \,\,\,\,\,\left| {{d_i} - {c_i}} \right| = {c_i} - {d_i} \leqslant {c_i} - {m_i} \leqslant \left| {{c_i} - {m_i}} \right| \leqslant {\varepsilon _i}\\ \end{gathered} \right\} \Rightarrow \,\left| {{d_i} - {c_i}} \right| \leqslant {\varepsilon _i} \;.$$
Therefore $$\sum\nolimits_i {\left| {{d_i} - {c_i}} \right|} \leqslant \sum\nolimits_i {{\varepsilon _i}} \leqslant \sum\nolimits_i {\left( {\left| {{a_i} - {c_i}} \right| + \left| {{b_i} - {c_i}} \right|} \right)} \leqslant 2L \;.$$
Is there any simpler proof?
For every $i$, $d_i-c_i=u_i(a_i-c_i)+v_i(b_i-c_i)$ hence $$|d_i-c_i|\leqslant u_i|a_i-c_i|+v_i|b_i-c_i|\leqslant |a_i-c_i|+|b_i-c_i|. $$ Summing this on $i$ yields $$ \sum\limits_i|d_i-c_i|\leqslant\sum\limits_i|a_i-c_i|+\sum\limits_i|b_i-c_i|\leqslant2L. $$