How to prove this $\frac{\sin{(nx)}}{\sin{x}}\ge\frac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$ let $n<\dfrac{\pi}{2\arccos{\dfrac{c}{2}}},c\in (0,2),c=2\cos{x}$, show that
$$\dfrac{\sin{(nx)}}{\sin{x}}\ge\dfrac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$$
where $0<x<\dfrac{\pi}{2}$
My idea: let $$a_{n}=\dfrac{\sin{(nx)}}{\sin{x}}$$
then for any $k\in[1,n]$, then we have
$$a^2_{k}-a^2_{k-1}=\dfrac{1}{\sin^2{x}}[\sin^2{kx}-\sin^2{(k-1)x}]=\dfrac{\sin{(2k-1)x}\sin{x}}{\sin^2{x}}=\dfrac{\sin{(2k-1)x}}{\sin{x}}$$
since
$$n<\dfrac{\pi}{2\arccos{\dfrac{c}{2}}}=\dfrac{\pi}{2x}\Longrightarrow 0<kx\le nx<\dfrac{\pi}{2}$$
then 
$2kx\le 2n<\pi$, so $x\le (2k-1)x<\pi-x$
so $\sin{(2k-1)x}\ge \sin{x}>0$,so
$a^2_{k}-a^2_{k-1}\ge 1$
$$a^2_{n}=\sum_{k=2}^{n}(a^2_{k}-a^2_{k-1})+a^2_{2}\ge n\Longrightarrow a_{n}\ge \sqrt{n}$$
then I can't it,
I'm sorry, I just to eat
,and I'm come back
 A: Next time,  I will to wait until the OP stops changing his/her question.
Note that
$$\frac{\sin nx}{\sin x}=U_n(\cos x)=2^{n-1}\prod_{k=1}^{n-1}\left(\cos x-\cos\left(\frac{k\pi}{n}\right)\right)$$
where $U_n$ is the Chebyshev polynomial of the second kind.
Now, for $0<x<\frac{\pi}{2n}$ (which is equivalent to the inequality with the $``c"$ thing,)
we have
$$\cos x-\cos\left(\frac{k\pi}{n}\right)\geq \cos\left( \frac{\pi}{2n}\right)-\cos\left(\frac{k\pi}{n}\right)>0$$
for $k=1,\ldots,n-1$. Thus
$$\frac{\sin nx}{\sin x}\geq2^{n-1}\prod_{k=1}^{n-1}\left(\cos \left(\frac{\pi}{2n}\right)-\cos\left(\frac{k\pi}{n}\right)\right)=\frac{\sin n(\frac{\pi}{2n})}{\sin (\frac{\pi}{2n})}=\frac{1}{ \sin (\frac{\pi}{2n})}.$$
This inequality is clearly better than the proposed one. In fact,
using $\sin (\frac{\pi}{2n})<\frac{\pi}{2n}$. The proposed inequality follows if we have
$$
\frac{2n}{\pi}\geq \frac{1}{\sqrt{3}}(2n-1)^{3/4}\iff n(2n)^3\geq \frac{\pi^4}{18}(2n-1)^3
$$
and this is clearly true if $n\geq \frac{\pi^4}{18}\approx 5.4$. So, the proposed inequality is true for $n\geq 6$. For $n\leq 5$ we can check directly that
$$\frac{1}{\sin(\frac{2n}{\pi})}\geq \frac{1}{\sqrt{3}}(2n-1)^{3/4}.$$
So, the proposed inequality is valid for every $n$. Note
that the inequality that we obtained is asymptotically much stronger.
A: This inequality $$\frac{\sin{(nx)}}{\sin{x}}\ge\frac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$$ with $0<x<\dfrac{\pi}{2}$ and $n>1$ makes a serious problem to me.
If we consider $$f_n(x)=\frac{\sin{(nx)}}{\sin{x}}-\frac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$$ we have $$f_n(0)=n-\frac{(2 n-1)^{3/4}}{\sqrt{3}} > 0$$ but it exists $0<x_n<\dfrac{\pi}{2}$ such that $f_n(x_n)=0$ and $x_n$ decreases when $n$ increases.
Built at $x=0$, a second order Taylor expansion gives $$x_n \simeq \frac{\sqrt{6 n-2 \sqrt{3} (2 n-1)^{3/4}}}{\sqrt{n \left(n^2-1\right)}}$$ which for large values of $n$ is $$x_n=\frac{\sqrt{6}}{n}-\sqrt[4]{2}
   \left(\frac{1}{n}\right)^{5/4}-\frac{\left(\frac{1}{n}\right)^{3/2}}{2
   \sqrt{3}}-\frac{\left(\frac{1}{n}\right)^{7/4}}{6 \sqrt[4]{2}}-\frac{5}{24
   \sqrt{6} n^2}+O\left(\left(\frac{1}{n}\right)^{9/4}\right)$$
Solving numerically $f_n(x)=0$ lead to $x_2=0.852587$, $x_3=0.543443$, $x_4=0.412797$, $x_5=0.337444$, $x_6=0.287336$, $x_7=0.251180$, $x_8=0.223690$, $x_9=0.201975$, $x_{10}=0.184339$ and above these values $f_n(x) < 0$.
