Find the conjugate function of $f(x)=\frac{\sum_{i=1}^{n}x_ie^{x_i/\gamma}}{\sum_{i=1}^{n}e^{x_i/\gamma}}$, where $x \in \mathbb{R}^n$.
By setting $p(x_i)=\frac{e^{x_i/\gamma}}{\sum_{j=1}^{n}e^{x_j/\gamma}}$, I write $f(x)=\sum_{i=1}^n x_i p(x_i)$.
Then, by finding the gradient of $y^Tx-f(x)$ and setting it to $0$, I have $y_i=\frac{x_i}{\gamma}p(x_i)(1-p(x_i))+p(x_i)$.
I write $x_i=\frac{\gamma(y_i-p(x_i))}{p(x_i)(1-p(x_i))}$ and subtitute it into $y^Tx-f(x)$ and get $f^*(y)=\sum_{i=1}^n\frac{\gamma(y_i-p(x_i))^2}{p(x_i)(1-p(x_i))}$.
But how can I eliminate $p(x_i)$ from the equation?