The first eigenvector of a distance matrix is nearly constant

Let $x_1,\dots ,x_n$ be points in a metric space and define the distance matrix $D\in {\mathbb{R}}^{n\times n}$ by $D_{ij}=d(x_i,x_j)$. The Perron-Frobenius Theorem implies that there is an eigenvector $v \in $\mathbb{R}$^n_{}$ with non-negative entries associated to the largest eigenvalue. We prove that this eigenvector is nearly constant in the sense that the inner product with the constant vector $\mathbb{1}\in {\mathbb{R}}^n$ is large $$⟨v,\mathbb{1}⟩\ge \frac{1}{\sqrt{2}}\cdot \mathrm{\parallel }v{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\cdot \mathrm{\parallel }\mathbb{1}{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}$$ and that each entry satisfies $v_i\ge \mathrm{\parallel }v{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\mathrm{/}\sqrt{4n}$. Both inequalities are sharp.