Suppose, $S$ is a convex set in $\mathbb{R^n}$and $ri(S)$ denotes the relative interior set of $S$.For, $x_1\in S$ and $x_2,x_3 \in ri(S)$ the following property holds
\begin{equation} d_{\phi}(x_1,x_3)=d_{\phi}(x_1,x_2)+d_{\phi}(x_2,x_3)-\langle(x_1-x_2)(\nabla \phi(x_3)-\nabla \phi(x_2))\rangle. \end{equation} where, $d_{\phi}(.,.)$ denotes the Bregman divergence with the convex function $\phi$.
Show that, when $x_1,x_2,x_3$ are such that $x_1 \in S'$, $S'$ being convex subset of $S$ and $x_2$ is given by
$x_2=\rm{argmin} ~\{d_{\phi}(x,x_3):x \in S'\}$ then the third term of the previous equation is negative.
Proof: when $x_2=\rm{argmin} ~\{d_{\phi}(x,x_3):x \in S'\}$ then,
$d_{\phi}(x_2,x_3)<d_{\phi}(x_1,x_3)$
Now, $\phi$ is a convex function,hence $(\nabla \phi(x_3)-\nabla \phi(x_2))>0$ whenever $x_3>x_2$. On the other hand,
$(x_1-x_2)>0$ if $x_1>x_2$.Therefore the third term is negative when both $x_1,x_3>x_2$.
But how can I prove it in general??Give me some hint please.