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

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The discussion focuses on proving the inequality involving the product of sine functions and their arguments, specifically that the product of the ratios of sine to the argument for a set of angles is less than or equal to the same ratio evaluated at the average of those angles. The angles are constrained to be between 0 and π. The proof hinges on properties of the sine function and inequalities related to convexity. Participants express appreciation for contributions and insights shared during the discussion. The overall goal is to establish a mathematical inequality that holds under the given conditions.
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Let $0<a_i<\pi$, $i=1,\,\cdots,\,n$ and let $a=\dfrac{a_1+\cdots+a_n}{n}$. Prove that $\displaystyle \prod_{i=1}^{n} \left(\dfrac{\sin a_i}{a_i}\right)\le \left(\dfrac{\sin a}{a}\right)^n$.
 
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anemone said:
Let $0<a_i<\pi$, $i=1,\,\cdots,\,n$ and let $a=\dfrac{a_1+\cdots+a_n}{n}$. Prove that $\displaystyle \prod_{i=1}^{n} \left(\dfrac{\sin a_i}{a_i}\right)\le \left(\dfrac{\sin a}{a}\right)^n$.

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

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