For some strictly convex functions ${f}_{1}({x}_{1}),{f}_{2}({x}_{2}),...,{f}_{n}({x}_{n})$, it seems intuitive that for a given sum of ${x}_{1}+{x}_{2}+...+{x}_{n}=X$, where ${x}_{1}$, ${x}_{2}$,..., ${x}_{n}\ge 0$

$\mathrm{\exists}k\in \{1,2,\dots ,n\},\phantom{\rule{thinmathspace}{0ex}}{f}_{k}(X)=max\{\sum _{i=1}^{n}{f}_{i}({x}_{i})\phantom{\rule{thinmathspace}{0ex}}:\sum _{i=1}^{n}{x}_{i}=X\text{and}\mathrm{\forall}i,\phantom{\rule{thinmathspace}{0ex}}{x}_{i}\ge 0\phantom{\rule{thinmathspace}{0ex}}\}$

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?

$\mathrm{\exists}k\in \{1,2,\dots ,n\},\phantom{\rule{thinmathspace}{0ex}}{f}_{k}(X)=max\{\sum _{i=1}^{n}{f}_{i}({x}_{i})\phantom{\rule{thinmathspace}{0ex}}:\sum _{i=1}^{n}{x}_{i}=X\text{and}\mathrm{\forall}i,\phantom{\rule{thinmathspace}{0ex}}{x}_{i}\ge 0\phantom{\rule{thinmathspace}{0ex}}\}$

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?