Generalized inequality

[SANZEED]

Let $A_{1},A_{2},...,A_{n}$ be angles belonging to the interval $[0,\pi]$. Prove that,
$\prod_{i=1}^{n}sin A_{i}\leq (sin \frac{\sum_{i=1}^{n}A_{i}}{n})$.

[SANZEED]

Here I have used Jensen.
We have,of course, $log (\Pi_{i=1}^{n}sin A_{i})=\sum_{i=1}^{n}log (sin A_{i})$. Since $log x$ is concave then we have $n log (\frac{\sum_{i=1}^{n}sin A_{i}}{n})\geq \sum_{i=1}^{n}log (sin A_{i})$. But again we have that since $sin x$ is concave on the given interval, $\frac{\sum_{i=1}^{n}sin A_{i}}{n}\leq sin \frac{\sum_{i=1}^{n}A_{i}}{n}$. From taking logarithms on both sides of the last inequality the desired result follows.

Apparently, this inequality can be used to solve IMO-$2001$-G$4$.

[Sazid Akhter Turzo]

Direct application of Jensen.
Use $\phi(x) = ln(sinx)$ that implies $\phi''(x) < 0$. So $\phi$ is concave.