The statement of the problem : Let $n \in \mathbb Z$ , $n \ge 3$ and consider the real numbers $ 0 \le x_1 \lt x_2 \lt x_3 \lt \ldots \lt x_n$ in arithmetic progression and also consider the real number $a \gt 1$. Find the smallest element of the set
$\mathcal S =\{ a^{x_i} + a^{x_j} + a^{x_k} \mid i,j,k \in \{1,2,3, \ldots,n \} , i+j+k=n \}$
My approach : I have been working on this problem for some time and I have no general idea, I tried with some smaller cases, like $n = 3$ we have the smallest element $a^{x_1} + a^{x_1} + a^{x_1}$. For $n = 4$ we have the smallest element $a^{x_1} + a^{x_1} + a^{x_2}$. For $n = 5$ answer could be either $a^{x_1} + a^{x_1} + a^{x_3}$ either $a^{x_1} + a^{x_2} + a^{x_2}$ because we would have to compare $2*a^{\frac{x_1+x_3}{2}}$ with $a^{x_1}+a^{x_3}$ , and using the inequality of means we get that $a^{x_1}+a^{x_3} >2*a^{\frac{x_1+x_3}{2}}$, so the smallest element is $a^{x_1} + a^{x_2} + a^{x_2}$.
I don't know how the general case should be approached, but I am open to your ideas and solutions. Thanks a lot!
So, in the case where $n=5$ you prove that $a^{x_1}+a^{x_2}+a^{x_2}\le a^{x_1}+a^{x_1}+a^{x_3}$ but then state that $a^{x_1}+a^{x_1}+a^{x_3}$ is minimum. No, $a^{x_1}+a^{x_2}+a^{x_2}$ is minimum as you just proved.
In general, the smallest element of the set $\{a ^{x_i}+a^{x_j}+a^{x_k}:i+j+k=n\}$ is obtained when exponents are (almost) equal. (Almost equal means that any two exponents are of the form $x_\alpha$, $x_\beta$, where $\alpha$ and $\beta$ are the same or differ by $1$). The easiest way to show that is to prove that if $$i\le j-2$$ then $$a^{x_i}+a^{x_j}+a^{x_k}>a^{x_{i+1}}+a^{x_{j-1}}+a^{x_k}.\tag1$$ Note that still $i+1\le j-1$. Applying this procedure several time, you will make exponents (almost) equal, and the value will get smaller every time. Let us prove $(1)$. It is equivalent to
$$a^{x_{j}} -a^{x_{j-1}}>a^{x_{i+1}}-a^{x_{i}}$$ $$a^{x_{j-1}+d} -a^{x_{j-1}}>a^{x_{i}+d}-a^{x_{i}},$$
where $d$ is the step of the arithmetic progression $\{x_\alpha\}$. Then:
$$a^{x_{j-1}}(a^d -1)>a^{x_{i}}(a^d-1).$$
Cancel $a^d-1>0$:
$$a^{x_{j-1}}>a^{x_{i}}.$$
This is true since $a>1$ and $j-1>i$.
There is another way to solve this. Let us assume, for the sake of simplicity, that $3\mid n$. Let us prove that for any $i,j,k$ such that $i+j+k=n$ holds $$a^{x_i}+ a^{x_j}+ a^{x_k} \ge 3a^{x_{n/3}}.$$
We will use AM-GM: $$a^{x_i}+ a^{x_j}+ a^{x_k} \ge 3(a^{x_i}a^{x_j}a^{x_k})^{1/3}= 3(a^{x_i+x_j+x_k})^{1/3}=$$
$$=3(a^{x_{n/3}+d\cdot(i-n/3)+ x_{n/3}+d\cdot(j-n/3)+ x_{n/3}+d\cdot(k-n/3)})^{1/3}=$$
$$=3a^{\frac13(3 x_{n/3} +d\cdot(i+j+k-n)) }=3a^{x_{n/3}}.$$
The equality holds if and only if $i=j=k=n/3$.