I am reading about 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.