Show that $ \sin\frac{\pi }{3n}\ge \frac{\sqrt3 }{2n} ,\quad \forall n\in\mathbb{N} $.

ytChen wrote: Show that $ \sin\frac{\pi }{3n}\ge \frac{\sqrt3 }{2n} ,\quad \forall n\in\mathbb{N} $. The same idea from the solution I gave can also be applied here. As a generalization, we have the following nice inequality: $$\sin\frac xn\ge\frac{\sin x}n$$for all positive integers $n$ and real numbers $0<x\le\tfrac\pi2$.

$\left|\sin\frac xn\right|\ge\frac{\left|\sin x\right|}n$, $\forall n\in\mathbb{N}$ and $\forall x\in\mathbb{R}$.

ytChen wrote: $\left|\sin\frac xn\right|\ge\frac{\left|\sin x\right|}n$, $\forall n\in\mathbb{N}$ and $\forall x\in\mathbb{R}$. With your constraints for $ n,x, $ consider $ \epsilon = \cos\frac{2x}{n} +i\sin \frac{2x}{n} . $ Then, $$ \epsilon^n -1 = (\epsilon - 1)\left( \epsilon^{n-1} +\epsilon^{n-2} +\cdots +1 \right)\implies $$$$\implies\left|\epsilon^n -1\right| = \left|\epsilon - 1\right|\cdot\left| \epsilon^{n-1} +\epsilon^{n-2} +\cdots +1 \right|\implies $$$$ \implies\sqrt{2-\cos 2x} \le n\cdot\sqrt{2-\cos \frac{2x}{n} }\implies \left|\sin\frac xn\right|\ge\frac{\left|\sin x\right|}n . $$

CatalinBordea wrote: ytChen wrote: $\left|\sin\frac xn\right|\ge\frac{\left|\sin x\right|}n$, $\forall n\in\mathbb{N}$ and $\forall x\in\mathbb{R}$. With your constraints for $ n,x, $ consider $ \epsilon = \cos\frac{2x}{n} +i\sin \frac{2x}{n} . $ Then, $$ \epsilon^n -1 = (\epsilon - 1)\left( \epsilon^{n-1} +\epsilon^{n-2} +\cdots +1 \right)\implies $$$$\implies\left|\epsilon^n -1\right| = \left|\epsilon - 1\right|\cdot\left| \epsilon^{n-1} +\epsilon^{n-2} +\cdots +1 \right|\implies $$$$ \implies\sqrt{2-\cos 2x} \le n\cdot\sqrt{2-\cos \frac{2x}{n} }\implies \left|\sin\frac xn\right|\ge\frac{\left|\sin x\right|}n . $$ very inteligent solution!

CatalinBordea wrote: ytChen wrote: $\left|\sin\frac xn\right|\ge\frac{\left|\sin x\right|}n$, $\forall n\in\mathbb{N}$ and $\forall x\in\mathbb{R}$. With your constraints for $ n,x, $ consider $ \epsilon = \cos\frac{2x}{n} +i\sin \frac{2x}{n} . $ Then, $$ \epsilon^n -1 = (\epsilon - 1)\left( \epsilon^{n-1} +\epsilon^{n-2} +\cdots +1 \right)\implies $$$$\implies\left|\epsilon^n -1\right| = \left|\epsilon - 1\right|\cdot\left| \epsilon^{n-1} +\epsilon^{n-2} +\cdots +1 \right|\implies $$$$ \implies\sqrt{2-\cos 2x} \le n\cdot\sqrt{2-\cos \frac{2x}{n} }\implies \left|\sin\frac xn\right|\ge\frac{\left|\sin x\right|}n . $$ Solution. It suffices to show $\left|\sin(ny)\right|\le n\left|\sin y\right|$ for every $n\in\mathbb{N}$ and every $y\in\mathbb{R}$. If $\left|\sin(ky)\right|\le k\left|\sin y\right|$, then \begin{align*}\left|\sin[(k+1)y]\right|=&\left|\sin(ky)\cos y+\sin y\cos(ky)\right|\\ \le&\left|\sin(ky)\right|\cdot\left|\cos y\right|+\left|\sin y\right|\cdot\left|\cos(ky)\right|\\ \le&\left|\sin(ky)\right|+\left|\sin y\right|\\ \le& (k+1)\left|\sin y\right|, \end{align*}which completes the induction. Taking $y=\frac{x}{n}$ leads to the desired inequality. $\blacksquare$

CatalinBordea wrote: $ \sin\frac{\pi }{4n}\ge \frac{\sqrt 2 }{2n} ,\quad \forall n\in\mathbb{N} $

https://artofproblemsolving.com/community/c6h1439207p8174608

Very neat solutions indeed.

I found a new solution Claim $\frac{sinx}{x} \ge \frac{2\sqrt{2}}{\pi} \forall x \in (0,\frac{\pi}{4})$. Proof Consider ,$f(x)=\frac{sinx}{x} $. $ or, f'(x)= \frac{xcosx-sinx}{x^2} $. $Or,f'(x)=\frac{cos x (x-tanx)}{x^2} $. $Or,f'(x) <0$. So ,f(x) is decreasing in interval $ (0,\frac{\pi}{4})$. So,$\frac{sinx}{x} \ge \frac{\frac{1}{\sqrt{2}}}{\frac{pi}{4}}$. Or,$\frac{sinx}{x} \ge \frac{2\sqrt{2}}{\pi} \forall x \in (0,\frac{\pi}{4})$. By LMVT I have done this . Now,put $ x= \frac{\pi}{4n}$.....get, $\boxed{ \sin\frac{\pi }{4n}\ge \frac{\sqrt 2 }{2n} ,\quad \forall n\in\mathbb{N} }$

ftheftics wrote: I found a new solution Claim $\frac{sinx}{x} \ge \frac{2\sqrt{2}}{\pi} \forall x \in (0,\frac{\pi}{4})$. Proof Consider ,$f(x)=\frac{sinx}{x} $. $ or, f'(x)= \frac{xcosx-sinx}{x^2} $. $Or,f'(x)=\frac{cos x (x-tanx)}{x^2} $. $Or,f'(x) <0$. So ,f(x) is decreasing in interval $ (0,\frac{\pi}{4})$. So,$\frac{sinx}{x} \ge \frac{\frac{1}{\sqrt{2}}}{\frac{pi}{4}}$. Or,$\frac{sinx}{x} \ge \frac{2\sqrt{2}}{\pi} \forall x \in (0,\frac{\pi}{4})$. By LMVT I have done this . Now,put $ x= \frac{\pi}{4n}$.....get, $\boxed{ \sin\frac{\pi }{4n}\ge \frac{\sqrt 2 }{2n} ,\quad \forall n\in\mathbb{N} }$ beautifull solution