Let $\mathcal{X}$ be a closed and convex set. Due to the non-expansive of projection as $$\|\text{Proj}_{\mathcal{X}}(x) - \text{Proj}_{\mathcal{X}}(y)\| \leq \|x-y\|,$$ where $\text{Proj}$ is the projection operator.
My question is that given $u, v \in \mathcal{X}$ and a constant $\alpha \in (0,1)$, for any $x, y$, will the following inequality hold? $$ \|(\alpha\text{Proj}_{\mathcal{X}}(x) + (1-\alpha)u) - (\alpha\text{Proj}_{\mathcal{X}}(y) + (1-\alpha)v\| \leq \|(\alpha x + (1-\alpha)u)-(\alpha y + (1-\alpha)v)\|.$$
Not necessarily. In $\Bbb{R}^2$, take \begin{align*} x &= (2, 2) \\ y &= (1, -1) \\ u &= (-2, -2) \\ v &= (-1, 1) \\ \alpha &= \frac{1}{2} \\ \mathcal{X} &= \operatorname{conv}\{u, v\}. \end{align*} Then, \begin{align*} \operatorname{Proj}_\mathcal{X}(x) &= v \\ \operatorname{Proj}_\mathcal{X}(y) &= \left(-\frac{7}{5}, -\frac{1}{5}\right), \end{align*} and $$\|(\alpha\operatorname{Proj}_\mathcal{X}(x) + (1 - \alpha)u) - (\alpha \operatorname{Proj}_\mathcal{X}(y) + (1 - \alpha)v)\| = \frac{3\sqrt{10}}{10},$$ (I think; it's definitely positive!) but $$\|(\alpha x + (1 - \alpha)u) - (\alpha y + (1 - \alpha)v)\| = 0.$$