I am seeking assistance in proving an inequality that I believe holds for a specific mathematical concept involving the proximal operator based on a proper convex function. The inequality is as follows: $$\sum_{i=1}^{n} \|\text{prox}_{h}(x_i) - \frac{1}{n} \sum_{j=1}^{n} \text{prox}_h(x_j)\|^2 \leq \sum_{i=1}^{n} \|x_i - \left(\frac{1}{n} \sum_{j=1}^{n} x_j\right)\|^2$$ with $\text{prox}_h(\cdot)$ the proximal operator based on proper convex function $h$.
This inequality is interesting because it relates to the comparison of two expressions involving proximal operators and serves as a potential tool for solving optimization problems. I think the non-expensive property, i.e., $\|\text{prox}_h(x) - \text{prox}_h(y)\| \leq \|x-y\|$ may serve as a key tool to prove it.
I have tried to prove it through several aspects, but find it difficult. A sufficient condition to prove it is to show that $$\sum_{i=1}^{n} \|\text{prox}_{h}(x_i) - \frac{1}{n} \sum_{j=1}^{n} \text{prox}_h(x_j)\|^2 \leq \sum_{i=1}^{n} \|\text{prox}_{h}(x_i) - \text{prox}_h(\frac{1}{n} \sum_{j=1}^{n} x_j)\|^2$$ which may hold since I have tried through random examples.
On the other hand, I have also tried to handle the average sum by using the convex property of the 2-norm: $$\sum_{i=1}^{n} \|\text{prox}_{h}(x_i) - \frac{1}{n} \sum_{j=1}^{n} \text{prox}_h(x_j)\|^2 \leq \sum_{i=1}^{n} \frac{1}{n} \sum_{j=1}^{n}\|\text{prox}_{h}(x_i) - \text{prox}_h( x_j)\|^2 \leq \sum_{i=1}^{n} \frac{1}{n} \sum_{j=1}^{n}\|x_i - x_j\|^2$$ However, it is hard to get back the average $\sum_{j=1}^{n} x_j $ in the norm.
I have solved this problem. We first introduce an auxiliary equation: For any vector $a_i \in \mathbb{R}^d$, it holds that $$\sum_{i}\sum_{j} \|a_i - a_j\|^2 = 2n \sum_{i} \|a_i - \bar{a}\|^2.$$
Then, we can convert the LHS of our inequality to $$\sum_{i=1}^{n} \|prox_{h}(x_i) - \frac{1}{n} \sum_{j=1}^{n} prox_h(x_j)\|^2 = \frac{1}{2n}\sum_{j=1}^{n}\sum_{k=1}^{n} \|prox_{h}(x_j) - prox_h(x_k)\|^2.$$ By the nonexpansivity of the proximal operator, we finally get $$ \frac{1}{2n}\sum_{j=1}^{n}\sum_{k=1}^{n} \|prox_{h}(x_j) - prox_h(x_k)\|^2 \leq \frac{1}{2n}\sum_{j=1}^{n}\sum_{k=1}^{n} \|x_j - x_k\|^2 = \sum_{i=1}^{n} \|x_i - \left(\frac{1}{n} \sum_{j=1}^{n} x_j\right)\|^2.$$