for a sine function, how to prove the follow inequality

Question

For a function $f(x)=\sin(x),x \in [0,\pi/2]$, We have a set of numbers $X=\{x_1,x_2,...,x_k\}$ and $\sum_{k} x_i =A\in [0,\pi/2]$, if we enlarge the size of set $X$ from $k$ to $k+l$ (keep the summation $A$ unchanged), and get a new set Y=$\{Y_1,Y_2\}$, where $Y_1=X/(\frac{k}{{k + l}})$ and $Y_2=\underbrace {\frac{1}{l}\frac{l}{{k + l}}A,\frac{1}{l}\frac{l}{{k + l}}A, \cdots ,\frac{1}{l}\frac{l}{{k + l}}A}_l$, please prove that $\sum_i f(x_i) < \sum_j f(y_j)$.

Answer 1

$f(x)=\sin(x)$ is a concave function on $\left[0,\frac{\pi}{2}\right]$, hence your inequality is a consequence of Karamata's inequality. For any $n\in[1,k+l]$ we may define the terms of two sequences as

$$\{a_n\}_{n=1}^{k+l} = \underbrace{0,0,0,\ldots,0}_{l\text{ times}},x_1,x_2,\ldots,x_k; $$ $$\{b_n\}_{n=1}^{k+l} = \frac{k}{k+l}x_1,\ldots,\frac{k}{k+l}x_k,\underbrace{\frac{A}{k+l},\ldots\frac{A}{k+l}}_{l\text{ times}}$$ and for any $n\in[1,k+l-1]$ we have $a_1+\ldots+a_n \leq b_1+\ldots+b_n$, while the sum of both sequences is $A$.