A function $f: \mathcal{X} \to \mathbb{R}$ is $\alpha$-strongly convex w.r.t. a norm $\|\cdot\|$ if $f(x) - f(y) \leq \langle \nabla f(x), x-y \rangle - \frac{\alpha}{2} \|x-y\|^2$, for all $x,y \in \mathcal{X}$.
Let $h: \mathcal{X} \to \mathbb{R}$ be a differentiable, 1-strongly convex function w.r.t. a norm $\|\cdot\|$. Define the Bregman divergence $\mathcal{B}_h: \mathcal{X} \times \text{int} \ \mathcal{X} \to \mathbb{R}$ w.r.t. $h$ as $\mathcal{B}_h(x,y) := h(x) - h(y) - \langle\nabla h(y),x-y\rangle$.
We can assume that for all $x \in \mathcal{X}$ and $y \in \text{int} \ \mathcal{X}$, it holds that $\mathcal{B}_h(x,y) \leq R_1$ and/or $\|x-y\| \leq R_2$, for $R_1,R_2>0$;
I would like to prove the following:
$$ \lambda \mathcal{B}_h(x,y) \leq \mathcal{B}_h(x,z) + \mathcal{B}_h(y,z), \tag{1} $$ for some constant $\lambda > 0$, for any $x \in \mathcal{X}$ and $y,z \in \text{int} \ \mathcal{X}$.
For the special case of $h(x) = \frac{1}{2}\|x\|^2_2$, we have $\mathcal{B}_h(x,y) = \frac{1}{2}\|x-y\|^2_2$ and $(1)$ holds for $\lambda = \frac{1}{2}$. However, to show it I used the fact that $\frac{1}{2}\|x-y\|^2_2 = \frac{1}{2}\|y-x\|^2_2$ (i.e. its symmetry), which is does not hold for general Bregman divergences.
Using the three-point equality $\mathcal{B}_h(x,y) = \mathcal{B}_h(x,z) + \mathcal{B}_h(z,y) - \langle \nabla h(y) - \nabla h(z), x- z \rangle$, the fact that $\frac{1}{2}\|x-z\|^2 \leq \mathcal{B}_h(x,z)$ and the Cauchy–Schwarz inequality, I was able to show that $$ \mathcal{B}_h(x,y) \leq 2\mathcal{B}_h(x,z) + \mathcal{B}_h(z,y) + \frac{1}{2}\|\nabla h(y) - \nabla h(z)\|^2_*. $$ Moreover, apparently $\frac{1}{2}\|\nabla h(y) - \nabla h(z)\|^2_* \leq \mathcal{B}_{h^*}(\nabla h(y),\nabla h(z)) = \mathcal{B}_h(z,y)$ (or $\mathcal{B}_h(y,z)$ by the symmetry of the dual norm $\|\cdot\|_*$), but even then, it does not seem to solve the problem.
Question: Is it possible to prove $(1)$? Is there a counter-example for (1)?
Here is a counterexample. Let $$f(x) = \exp(x) + \frac12 x^2.$$ Choose $x = z = 0$ and $y$ large. Then, \begin{align*} B_f(x,y) &\approx y \exp(y), \\ B_f(x,z) &= 0, \\ B_f(y,z) &\approx \exp(y). \end{align*} Thus, you need $\lambda \le 1/y$ but this is arbitrarily close to $0$.