Given $\pmb X \in \mathcal{R}^p$, denote the elements of $\pmb X$ as $\pmb x_i$ for $i= 1, \dots, n$. Denote the $t(\pmb X)$ as the average of $\pmb X$ \begin{equation} \pmb t(\pmb X) = \frac 1 n \sum_{i=1}^n \pmb x_i \end{equation}
so, $\pmb t(\pmb X)$ is a vector and I can takes it Euclidean norm as $\|\pmb t(\pmb X)\|$. Further, denote $\pmb x_m$ where $m$ refers to the $m$th index, such that $\| \pmb x_m\| \ge \| \pmb x_i\|$ for any $i \ne m$. In other words, $\pmb x_m$ has the maximum norm of any element in $\pmb X$.
My question is this. Is it true that \begin{equation} \|\pmb t(\pmb X)\| \le \|\pmb x_m\|\text{ ?} \end{equation} This seems obvious to me, but I would like to confirm that it is obvious enough to claim without a proof. If a proof is necessary, is Jensen's inequality a way to show this, and if so how (to my disgrace, clearly the fact that I am not sure indicates a proof is necessary at least for me)?
By homogeneity and the triangle inequality $$\left\|\frac{1}{n}\sum_{k=1}^nx_k \right\|=\frac{1}{n}\left\|\sum_{k=1}^nx_k \right\| \leq \frac{1}{n}\sum_{k=1}^n\|x_k \| \leq\frac{1}{n}\sum_{k=1}^n\|x_m \| =\|x_m\| $$