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In a complete smooth simply connected Riemannian manifold of non-positive curvature, the square of the distance function $d^2(p,x)$ is a smooth strictly convex function of $x$. It follows that this is also true for positive linear combinations. What happens with affine combinations, namely combinations of the form $\sum w_i d^2(p_i,x)$, where $\sum w_i =1$? In euclidean space any affine combination of squared distance functions is strictly convex.

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First of all, you want non-positive, not non-negative curvature. In positively curved spaces $d^2$ is neither smooth nor convex (consider the sphere).

The answer is negative. Consider a geodesic $\gamma:[0,1]\to M$, and on it the affine combination $$f(t) = 2d^2(\gamma(t),\gamma(0))-d^2(\gamma(t),p) \tag{1}$$ where $p$ is some point in $M$. Since $\gamma$ is a geodesic, (1) simplifies to $$f(t) = 2t^2 - d^2(\gamma(t),p) \tag{2}$$ So now you are asking for $d^2(\gamma(t),p)-2t^2$ to be concave, which has the meaning of a lower curvature bound. In any sufficiently negatively curved space this property will fail.

Instead of justifying the previous sentence, I'll give a concrete example. Consider the Poincaré disk with the metric $$d(a,b) = \tanh^{-1}\left|\frac{a-b}{1-a\bar b}\right|$$ Consider the geodesic $\gamma(t) = \tanh t$ and the point $p=0.9i$. The function (2) takes the form $$2t^2 - \left(\tanh^{-1}\left|\frac{\tanh t-0.9i}{1+0.9 i \tanh t}\right| \right)^{2}$$ which is shown below:

counterexample

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