 # For some strictly convex functions f 1 </msub> ( x 1 </msub> kromo8hdcd 2022-05-09 Answered
For some strictly convex functions ${f}_{1}\left({x}_{1}\right),{f}_{2}\left({x}_{2}\right),...,{f}_{n}\left({x}_{n}\right)$, it seems intuitive that for a given sum of ${x}_{1}+{x}_{2}+...+{x}_{n}=X$, where ${x}_{1}$, ${x}_{2}$,..., ${x}_{n}\ge 0$

Since these functions are convex, the sum of these functions is greatest when the sum of arguments is equal to the argument of just one function. This seems intuitive, but I can't think of how to argue this in formal maths. Any suggestions?
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You can use the following theorem: if the maximum of a convex function on a convex set is attained, then it is attained at an extreme point.

In your case, the convex set is a simplex, which is a compact set. The function $F\left(x\right)=\sum _{i=1}^{n}{f}_{i}\left({x}_{i}\right)$ is convex, and thus continuous. Therefore, the maximum is attained by Weirstrass. The constraints define a convex polytope, whose extreme points are its vertices, which are . The result you wish follows from the fact that the maximum is attained at one of those vertices.