I am reading robust optimization and there is a claim:

$$ \max_{\mu^-\leq\mu\leq \mu^+} \ln \left( \exp\left\{w+\ln\left(\frac{1+\mu}{2}\right)\right\} +\exp\left\{-w+\ln\left(\frac{1-\mu}{2}\right)\right\} \right) $$ is equivalent to $$ \max_{-1\leq u\leq1} \{wu-\phi(u)\}, $$ with $$ \phi(u)= \begin{cases} \frac{1}{2}[(1+u)\ln\left(\frac{1+u}{1+\mu^-}\right) +(1-u)\ln\left(\frac{1-u}{1-\mu^-}\right)], & -1\leq u\leq\mu^-, \\ 0, & \mu^-\leq u \leq \mu^+, \\ \frac{1}{2}[(1+u)\ln\left(\frac{1+u}{1+\mu^+}\right) +(1-u)\ln\left(\frac{1-u}{1-\mu^+}\right)], & \mu^+\leq u\leq1. \end{cases} $$ when $-1\leq \mu^-\leq\mu^+\leq1$.

Could anyone help me prove this? There is a hint using the equality $$ \ln(\exp(x_1)+...+\exp(x_n))=\max_y\left\{x^Ty-\sum_{i=1}^{n}y_i\ln y_i:y\geq0,\sum_i y_i=1\right\}. $$ But I don't know how to use this equality