Let $0<\alpha_1 \leq \cdots \leq\alpha_m \leq \beta_1 \leq \cdots \leq \beta_n < \pi$ such that $\sum_{i=1}^m \alpha_i = \sum_{j=1}^n \beta_j = 2\pi.$
Can we show that $\sum_{i=1}^m \sin (\alpha_i/2) \geq \sum_{j=1}^n \sin (\beta_j/2)$?
More generally, is the following true? It looks like an improvement of Karamata's inequality in a way ($n\neq m$).
Let a convex fonction $ f:\mathbb{R} \to \mathbb{R}$ and $0<\alpha_1 \leq \cdots \leq\alpha_m \leq \beta_1 \leq \cdots \leq \beta_n < 1$ such that $\sum_{i=1}^m \alpha_i = \sum_{j=1}^n \beta_j=1 $. Then $\sum_{i=1}^m f(\alpha_i) \leq \sum_{j=1}^n \beta_j=1 $.
In the first case, note that $m> n$ and we may "pad" $m-n$ zeroes to $\{\beta_k\}$ without changing the inequality to prove. Then as $(\underbrace{0, 0, ...0}_{m-n \text{ terms}}, \beta_1, \beta_2, ... \beta_n) \succ (\alpha_1, \alpha_2, ... \alpha_m) $ and $t \mapsto \sin t/2$ is concave, we have the result using the traditional Karamata inequality.
In the more general case you suggest, the result does not hold. For instance, take $\alpha_k = 1/m$ and $\beta_k = 1/n$, then for the function $f(x)=1/x$ which is convex, the inequality reads $m^2< n^2$ which is obviously false.