Proof of decreasing distance of a simple function on a simplex $\newcommand {\ssum}{{\scriptscriptstyle \Sigma}} $
This problem arises in studying the limiting properties of a two-level hierarchical Gaussian Bayes model.  $ \sigma $ is the standard deviation of the variation of the unknown $x$ between samples. $s_i$ is the standard error if the measurement in sample $i$.  The $w_i$ are the weights used in a weighted average of the sample measurements $x_i$.
Define the L1-normed coordinates $\hat{w}_i(s) = w_i(s)/w_\ssum(s): 1 \le i \le N$, where
$$ w_i(s) = 1/(s_i+s)$$
$$w_\Sigma(s) = \sum_{i=1}^N w_i(s)$$
$$s_i>0 \;\; \forall i$$
$$ \sum_{i=1}^N \hat{w}_i(s) = 1$$
The $\{\hat{w}_i(s)\}$ specify a point on a simplex.
The Euclidean squared-distance from the origin is $D^2(s) =\sum_{i=1}^N \hat{w}_i^2(s)$.  $D^2(0)$ has a maximum of 1 when $s_i= c \,\delta_{ij}$ for some $j$ (i.e., only one $s$ is non-zero). When $s \to \infty$, all $\hat{w}_i(s) \to 1/N$ and $\lim_{s\to \infty}D^2(s)= 1/N$. I believe (but cannot prove) that this is a minimum.
I believe (but cannot prove) that $d D^2(s)/ds <0$ everywhere.  How can I do this?
 A: $\newcommand{\beps}{{\pmb \varepsilon}}$
The proof is mostly geometric:
Define orthonormal basis vectors in Euclidean $N$-space as $\beps^i\; (1 \le i \le N)$. The points specified by the $L_1$-normed vectors  $\check{\pmb W}= \sum_{i=1}^N \check{w}_i \,\beps^i$, where $\check{w}_i = \frac{w_i}{\sum_j w_j}$, lie in a standard $(N-1)$-simplex. The points specified by the $L_2$-normed vectors $\tilde{\pmb W} = \sum_{i=1}^N \tilde{w}_i \, \beps^i$,  where $\tilde{w}_i = \frac{w_i}{\sqrt{\sum_j w_j^2}}$, lie in the unit hypersphere, which intersects the vertices of the simplex. Specify indices so that $w_1$ is the largest $w_i$ (or one of the largest, if the maximum is not unique). The points $\beps^1$, $\tilde{\pmb W}$, and the origin $\pmb O$ define an isosceles triangle with base $\tilde{\pmb W}- {\beps^1}$ having squared-length $$||\tilde{\pmb W}- \beps^1||^2 = (1-\tilde{w}_1)^2 + 1- \tilde{w}_1^2= 2(1-\tilde{w}_1)\,.$$
The opposing vertex angle is
$$\theta = 2 \, \arcsin(\sqrt{1-\tilde{w}_1}/2)\,.$$
The vectors $\check{\pmb W}$ and $\check{w}_1 \, \beps^1$ are coplanar with the isosceles triangle.  The points $\pmb O$, $\check{w}_1 \, \beps^1$, and $\check{\pmb W}$ are the vertices of a right-triangle with hypotenuse $\check{\pmb W}$. The angle at the origin is $\theta$, so
$$||\check{\pmb W}|| = D(s)=  \check{w}_1/\cos \theta \,.$$
It is easy to prove that $d \check{w}_1/d s^2 <0$ and  $d \tilde{w}_1/d s^2 <0$. This implies that $d \theta/d s^2 >0$ and, then, that  $d ||{\pmb W}||/d s^2 <0$.    Q.E.D.
By the definition of the $w_i(s)$, $w_1(s)$ remains the largest (or one of the largest) $w_i(s)$ as $s$ increases. Therefore, $w_1(x) >1/N$ for finite $s$; otherwise there would be a $w_i(s)>w_1(s)$. Since it keeps decreasing $\lim_{s \to \infty}w_1(s) = 1/N$.  The remaining $w_i(\infty)= 1/N$ of $w_i(\infty) > w_1(\infty)$.  Therefore $\lim_{s \to \infty} D^2(s) = 1/N$.
