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.

  • 2
    $\begingroup$ Use $\sin (\alpha+\beta)$ formula. $\endgroup$
    – Aleksandar
    Jul 25, 2015 at 2:31
  • $\begingroup$ Can you show me the working steps using the same? $\endgroup$ Jul 25, 2015 at 2:33
  • $\begingroup$ Try to prove $n=2$ by using trigonometric sum to product identity, then use induction to prove if it's true for $n=2^k$, it's true for $n=2^{k+1}$. For $n\neq 2^k$, look at my answer of the following question math.stackexchange.com/questions/939869/… for the hint. $\endgroup$
    – user175968
    Jul 25, 2015 at 2:38
  • $\begingroup$ @frank000 but if it is true for $2^3=8$ you then only proof that it is true for $2^4=16$, so is it true for $11$? $\endgroup$ Jul 25, 2015 at 3:58
  • 1
    $\begingroup$ The key is to recognize that the sine function is concave on $[0,\pi]$. I used this property in my answer. Please let me know how I can improve my answer. I just want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Jul 25, 2015 at 5:54

1 Answer 1


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.

  • $\begingroup$ Is it that simple ;) Vote up! $\endgroup$ Jul 25, 2015 at 5:57
  • 1
    $\begingroup$ @johannesvalks Yes. In fact, we don't need induction unless we think of recursion as induction. $\endgroup$
    – Mark Viola
    Jul 25, 2015 at 5:58

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