The construction of \( \mu^{\prime}_{j} \) is such that \( \mu^{\prime}_{j} (x) \ge \mu_{j} (x) \) for all x, because d(x, b j ) ≤ d(x, c r*,j ) and thus \( {\frac{{d(x,b_{j} )}}{{d(x,c^{r*,t} )}}} \le {\frac{{d(x,c^{r*,j} )}}{{d(x,c^{r*,t} )}}} \) for all t. From this, it is also clear that \( \left( {{\frac{{d(x,b_{j} )}}{{d(x,c^{r*,t} )}}}} \right)^{{{2 \mathord{\left/ {\vphantom {2 {(m - 1)}}} \right. \kern-\nulldelimiterspace} {(m - 1)}}}} \le \left( {{\frac{{d(x,c^{r*,j} )}}{{d(x,c^{r*,t} )}}}} \right)^{{{2 \mathord{\left/ {\vphantom {2 {(m - 1)}}} \right. \kern-\nulldelimiterspace} {(m - 1)}}}} \) and \( 1 + \sum\nolimits_{t \ne j} {\left( {{\frac{{d(x,b_{j} )}}{{d(x,c^{r*,t} )}}}} \right)^{{{2 \mathord{\left/ {\vphantom {2 {(m - 1)}}} \right. \kern-\nulldelimiterspace} {(m - 1)}}}} } \le1 + \sum\nolimits_{t \ne j} {\left( {{\frac{{d(x,c^{r*,j} )}}{{d(x,c^{r*,t} )}}}} \right)^{{{2 \mathord{\left/ {\vphantom {2 {(m - 1)}}} \right. \kern-\nulldelimiterspace} {(m - 1)}}}} .} \) Therefore, \( \left( {1 + \sum\nolimits_{t \ne j} {\left( {{\frac{{d(x,b_{j} )}}{{d(x,c^{r*,t} )}}}} \right)^{{{2 \mathord{\left/ {\vphantom {2 {(m - 1)}}} \right. \kern-\nulldelimiterspace} {(m - 1)}}}} } } \right)^{ - 1} \ge \left( {1 +\sum\nolimits_{t \ne j} \left( {{\frac{{d(x,c^{r*,j} )}}{{d(x,c^{r*,t} )}}}} \right)^{{{2 \mathord{\left/ {\vphantom {2 {(m - 1)}}} \right. \kern-\nulldelimiterspace} {(m - 1)}}}} } \right)^{ - 1} \) and, thus, \( \mu^{\prime}_{j} (x) \ge \mu_{j} (x) \). Note that d(x, b j ) ≤ d(x, c r*,j) by construction of b j .