Prove $\prod\limits_{i=1}^{n}\frac{\sin a_i}{a_i}\le(\frac{\sin a}{a})^n$

[anemone]

Let $0<a_i<\pi$, $i=1,\,\cdots,\,n$ and let $a=\dfrac{a_1+\cdots+a_n}{n}$. Prove that $\displaystyle \prod_{i=1}^{n} \left(\dfrac{\sin a_i}{a_i}\right)\le \left(\dfrac{\sin a}{a}\right)^n$.

 

[Euge]

Let $0<a_i<\pi$, $i=1,\,\cdots,\,n$ and let $a=\dfrac{a_1+\cdots+a_n}{n}$. Prove that $\displaystyle \prod_{i=1}^{n} \left(\dfrac{\sin a_i}{a_i}\right)\le \left(\dfrac{\sin a}{a}\right)^n$.

Spoiler

Since the natural log function is concave on $(0, \infty)$ and the sine function is positive and concave on $(0, \pi)$, the composition $x \mapsto \ln \sin x$ is concave on $(0, \infty)$. Therefore

$\displaystyle \frac{1}{n}\sum_{i = 1}^n \ln \sin a_i \le \ln \sin a$.

Subtracting $a$ from both sides and using the formula $a = \frac{a_1 + \cdots + a_n}{n}$ results in

$\displaystyle \frac{1}{n} \sum_{i = 1}^n (\ln \sin a_i - a_i) \le \ln \sin a - a$,

or

$\displaystyle \frac{1}{n} \sum_{i = 1}^n \ln \frac{\sin a_i}{a_i} \le \ln \frac{\sin a}{a}$.

Multiplying by $n$ and exponentiating yields

$\displaystyle \prod_{i = 1}^n \frac{\sin a_i}{a_i} \le \left(\frac{\sin a}{a}\right)^n$.

 

[anemone]

Spoiler

Since the natural log function is concave on $(0, \infty)$ and the sine function is positive and concave on $(0, \pi)$, the composition $x \mapsto \ln \sin x$ is concave on $(0, \infty)$. Therefore

$\displaystyle \frac{1}{n}\sum_{i = 1}^n \ln \sin a_i \le \ln \sin a$.

Subtracting $a$ from both sides and using the formula $a = \frac{a_1 + \cdots + a_n}{n}$ results in

$\displaystyle \frac{1}{n} \sum_{i = 1}^n (\ln \sin a_i - a_i) \le \ln \sin a - a$,

or

$\displaystyle \frac{1}{n} \sum_{i = 1}^n \ln \frac{\sin a_i}{a_i} \le \ln \frac{\sin a}{a}$.

Multiplying by $n$ and exponentiating yields

$\displaystyle \prod_{i = 1}^n \frac{\sin a_i}{a_i} \le \left(\frac{\sin a}{a}\right)^n$.

Well done, Euge! And thanks for participating!:)