Let $\textbf{v} = (v_1, v_2, \dots, v_n) \in \mathbb{R}^n_{\geq 0}$ and $\textbf{w} = (w_1, w_2, \dots, w_n) \in \mathbb{R}^n_{\geq 0}$ be such that each $v_i \geq 0$ and each $w_i \geq 0$. Moreover, $\textbf{v} \neq \textbf{w}$. Define the "arithmetic mean" of $\textbf{v}$ and $\textbf{w}$ by
$$\textbf{a} = \left( \frac{v_1 + w_1}{2}, \frac{v_2 + w_2}{2}, \dots, \frac{v_n + w_n}{2} \right) \in \mathbb{R}^n_{\geq 0}.$$
For lack of a better term, define the "quadratic mean" (or "root mean square") of $\textbf{v}$ and $\textbf{w}$ by
$$\textbf{q} = \left( \sqrt{\frac{v_1^2 + w_1^2}{2}}, \sqrt{\frac{v_2^2 + w_2^2}{2}}, \dots, \sqrt{\frac{v_n^2 + w_n^2}{2}} \right) \in \mathbb{R}^n_{\geq 0}.$$
I want to show that q is closer to a than v is. Specifically, with
$$d(\textbf{x},\textbf{y}) = \sqrt{\sum_{i = 1}^n (x_i - y_i)^2}$$
denoting the Euclidean distance between $\textbf{x} \in \mathbb{R}^n$ and $\textbf{y} \in \mathbb{R}^n$, I want to show that
$$d(\textbf{q},\textbf{a}) < d(\textbf{v},\textbf{a}).$$
I have tried millions of random $\textbf{v}$ and $\textbf{w}$ without finding a counter-example, so I am quite sure that this is true. Any help on proving this is appreciated!