I was solving the following problem: $$n^{a_1}+n^{a_2}+...+n^{a_{n+1}}\geq n^{a_1^{\frac{1}{n}}}+n^{a_2^{\frac{1}{n}}}+...+n^{a_{n+1}^{\frac{1}{n}}}$$ Where $n$ was meant to be natural, the $a_i$ positive and the constraint $\prod_{k=1}^{n+1}a_k=1$ was given. Let $a_i\geq a_j\iff i \geq j$.
If $n=1$, the problem statement is trivial, so let $n\geq 1$. The way I now attempted to solve this was using the function $f(t)=e^{\ln(n)e^t}$. By manual verification, this function is convex on $\mathbb{R}$. Let $t_i=\ln(a_i)$ Then the problem's constraint becomes:
$$\sum_{k=1}^n t_i=0$$
The problem now becomes: $$f(t_1)+f(t_2)+ \dots +f(t_{n+1})\geq f\left( \frac{t_1}{n} \right) + f\left( \frac{t_2}{n} \right) + \dots + f\left( \frac{t_{n+1}}{n} \right)$$ But by construction, the $t_i$ majorize the $\frac{t_i}{n}$, so by Karamata, the inequality holds. Is this proof correct?
Either way, I would be very curious to see a non-analytic proof.
Without Karamata:
By Bernoulli, we have $a_1^{1/n} \le 1 + (a_1 - 1)/n$. It suffices to prove that $$n^{a_1} + \cdots + n^{a_{n+1}} \ge n^{1 + (a_1 - 1)/n} + \cdots + n^{1 + (a_{n+1} - 1)/n}$$ or $$n^{a_1-1} + \cdots + n^{a_{n+1} - 1} \ge n^{(a_1 - 1)/n} + \cdots + n^{(a_{n+1} - 1)/n}$$ or $$b_1^n + b_2^n + \cdots + b_{n+1}^n \ge b_1 + b_2 + \cdots + b_{n+1}$$ where $b_k = n^{(a_k - 1)/n}, ~ k=1, \cdots, n+1$.
By Bernoulli, we have $b_1^n \ge 1 + (b_1 - 1)n$. It suffices to prove that $$(n+1) + (b_1 + \cdots + b_{n+1} - (n+1))n \ge b_1 + \cdots + b_{n+1}$$ or $$b_1 + \cdots + b_{n+ 1} \ge n + 1$$ which is true, using $$b_1 b_2 \cdots b_{n+1} = n^{(a_1 + a_2 + \cdots + a_{n+1} - (n+1))/n} \ge 1.$$
We are done.