Let $0<A_i<\pi$ for $i=1,2,3,\ldots,n$. Use mathematical induction to prove that $$\sin A_1+\sin A_2+\cdots+\sin A_n\le n \sin\left(\frac{A_1+A_2+A_3+\cdots+A_n} n\right)$$ where $n\ge 1$ is a natural number.`
The inequality holds true for $n=1$. I assumed that it holds true for $n=k$. But I was unable to prove that it holds true for $n=k+1$. Please help me.
Recall that the sine function is Concave on $[0,\pi]$ so that for any points $x$ and $y$ on $[0,\pi]$ and any value of $t\in(0,1)$, we have
$$\sin(tx+(1-t)y)\ge t\sin x+(1-t)\sin y \tag 1$$
Now, letting $x=\frac1n\sum_{i=1}^{n}A_i$, $y=A_{n+1}$, $t=\frac{n}{n+1}$ and $1-t=\frac{1}{n+1}$ in $(1)$, we have
$$\begin{align} \sin\left(\frac{1}{n+1}\sum_{i=1}^{n+1}A_i\right)&=\sin\left(\frac{n}{n+1}\left(\frac1n\sum_{i=1}^{n}A_i\right)+\frac{1}{n+1}A_{n+1}\right)\\\\ &\ge \frac{n}{n+1}\sin\left(\frac1n\sum_{i=1}^{n}A_i\right)+\frac{1}{n+1}\sin\left(A_{n+1}\right) \tag 2 \end{align}$$
It is key to note that $\frac{1}{n}\sum_{i=1}^{n}A_i\in[0,\pi]$.
Next, we could apply the inequality in $(2)$ to the sum on the right-hand side of $(2)$ and through a recursive process arrive at the desired inequality.
But the OP specifically asked to show that the inequality of interest through induction. So, to that end, we establish a base case. Using $(1)$ with $t=1/2$, we see that
$$2\sin \left(\frac{A_1+A_2}{2}\right)\ge \sin A_1+\sin A_2$$
Next, we assume that for some integer $n$, we have
$$n\,\sin\left(\frac1n\sum_{i=1}^{n}A_i\right)\ge \sum_{i=1}^{n}\sin(A_i) \tag 3$$
Using $(3)$ in $(2)$ reveals
$$\begin{align} \sin\left(\frac{1}{n+1}\sum_{i=1}^{n+1}A_i\right)&\ge \frac{n}{n+1}\left(\frac1n \sum_{i=1}^{n}\sin(A_i)\right)+\frac{1}{n+1}\sin\left(A_{n+1}\right)\\\\ &=\frac{1}{n+1}\sum_{i=1}^{n+1}\sin(A_i) \tag 4 \end{align}$$
where after multiplying $(4)$ by $n+1$ yields the desire result.