$\exp(a)+\exp(b)+\exp(c)\leq \exp(1)+\exp(0)+\exp(0)$ when $0\leq a,b,c\leq 1$ and $a+b+c=1$

It seems that:

Under the constraint of:

$x_1+x_2+...+x_n=1, 0\le x_1,x_2,...,x_n \le 1$,

We have:

$e^{x_1}+e^{x_2}+...+e^{x_n} \le e^1+e^0+...+e^0$

And it is equal only when one of the $x$ reaches $1$ and others remain $0$.

Is there mathematical proof to explain it?

The function $$(x_1, \ldots x_n) \mapsto e^{x_1} + \ldots + e^{x_n}$$ on $\mathbb R^n$ is a sum of convex functions and therefore convex. So on any compact convex subset of $\mathbb R^n$ it takes its maximum in an extreme point. For the convex hull of the standard basis vectors it is therefore maximal in a basis vector (and, by symmetry, in all basis vectors).

Hint :

$e^x$ is an convex function ( for all $x$ and in particular for $x\geq 0$ ), and the vector $(1,0\cdots,0)$ majorizes the vector $(x_1,...,x_n)$ .So we can apply Karamata's inequality