<s>[INST] Prove: $\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N}\setminus\{0\}, x\in{\mathbb{R}, x\neq k\pi}} $ 
Prove :
$\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N\setminus\{0\}}}, x\in{\mathbb{R}} $

I proved this relationship by incident. I tried to directly prove this afterwards, but failed. I would love to see another proof to this Problem. 
-A proof-: we know that $\displaystyle\sum_{i=1}^{n}\cos{a_i}= \sum_{k=1}^{n}\cos{(a+(k-1)x)}=\frac{\cos{\frac{a+a_n}{2}}\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}=T(x)\quad(1)$
(Here's the source for $(1)$).
Also, $\displaystyle\sum_{i=1}^{n}a^2_i=\sum_{k=1}^{n}(a+(k-1)x)^2\\=\displaystyle\sum_{k=1}^{n}(a^2+2ax(k-1)+x^2(k-1)^2)\\=\displaystyle\sum_{k=1}^{n}a^2+2ax\sum_{k=1}^{n}(k-1)+x^2\sum_{k=1}^{n}(k-1)^2\\=na^2+2ax\frac{n(n-1)}{2}+x^2\frac{n(n-1)(2n-1)}{6}\quad(2)$
We consider the function $f(x)=\frac{x^2}{2}+\cos{x}\implies f''(x)=1-\cos{x}\geq{0}$ therefore $f(x)$ is a concave function. From Jensens inequality for concave functions:$ \displaystyle\sum_{i=1}^{n}f(a_i)\geq{nf\left(\frac{\sum_{i=1}^{n}a_i}{n}\right)}\iff\displaystyle\sum_{i=1}^{n}\left(\frac{a^2_i}{2}+\cos{a_i}\right)\geq n\Big(\frac{1}{2}\left(\frac{\frac{n}{2}(a+a_n)}{n}\right)^2+\cos{\frac{\frac{n}{2}(a+a_n)}{n}}\Big)\iff\frac{1}{2}\sum_{i=1}^{n}a^2_i+\sum_{i=1}^{n}\cos{a_i}\geq \frac{n}{2}\left(\frac{a+a_n}{2}\right)^2+n\cos{\frac{a+a_n}{2}}\overset{(1),(2)}{\iff}\frac{1}{2}\left(na^2+2ax\frac{n(n-1)}{2}+x^2\frac{n(n-1)(2n-1)}{6}\right)+T(x)\geqslant \frac{n\left(2a+(n-1)x\right)^2}{8}+n\cos{\frac{a+a_n}{2}}\overset{\ldots}{\iff} \frac{x^2n(n^2-1)}{3}\geqslant n\cos{\frac{a+a_n}{2}}-T(x)\overset{(1)}{\iff} \frac{x^2n(n^2-1)}{3}\geq \cos{\frac{a+a_n}{2}}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\iff \frac{x^2n(n^2-1)}{3}\geqslant \cos{\left(a+\frac{(n-1)x}{2}\right)}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\quad(3) $
$Lemma.$ For every dinstict $x,x\in\mathbb{R}_{\neq kπ}$ there exists at least one value of $a$ such that $\cos{(a+\frac{(n-1)x}{2}})=1\quad(5)$
Proof of the lemma: $(5)\iff (n-1)x=2-2a+4kπ\iff a=2kπ+1-\frac{(n-1)x}{2}$.
Hence, for every $x,x\in\mathbb{R}_{\neq k\pi}$
we have  $\frac{x^2n(n^2-1)}{3}\geqslant \cos{\left(a+\frac{(n-1)x}{2}\right)}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\overset{(4,x\to 2x)}{\iff}\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}} \square$


*

*To me it looks like $n-\frac{x^2n(n^2-1)}{6}$ is a good approximation function of $\frac{\sin{nx}}{\sin{x}}$ for small values of $x$ (or for small values of $x+ z\pi$, $z\in\mathbb{Z}$, it depends on $n$, because the second one is periodic with period multiple of $\pi$).



 [/INST] We will prove a more general result. 
Let
$$
f(x)=\frac{\sin nx}{\sin x}=\sum_{k=0}^\infty a_kx^{2k}\tag1
$$
and 
$$
f_m(x)=\sum_{k=0}^m a_kx^{2k}.\tag2
$$
Then the following inequalities hold:
$$\begin{cases}
f(x)\le f_m(x),& m\text{ even}\\
f(x)\ge f_m(x),& m\text{ odd}\\
\end{cases}.\tag3
$$
Your inequality will then follow as a special case: $m=1$.
The first key point of the proof is the observation:
$$
\frac{\sin nx}{\sin x}=\sum_{\ell=0}^{n-1}e^{i(n-1-2\ell)x}
=\begin{cases}
1+2\sum_{\ell=1}^{\frac{n-1}2}\cos2\ell x,&n\text{ odd}\\
2\sum_{\ell=1}^{\frac{n}2}\cos(2\ell-1)x,&n\text{ even},\tag4
\end{cases}
$$
which follows from simple telescopic identity:
$$
(e^{ix}-e^{-ix})\sum_{\ell=0}^{n-1}e^{i(n-1-2\ell)x}=e^{inx}-e^{-inx}.
$$
The second key point is the well-known inequalities (which can be proved e.g. by $2m$-fold integration of the inequality  $1-\cos x\ge0$):
$$
\begin{cases}
\cos x\le \sum_{k=0}^m c_kx^{2k},& m\text{ even}\\
\cos x\ge \sum_{k=0}^m c_kx^{2k},& m\text{ odd},
\end{cases}\tag5
$$
where $c_k=\frac{(-1)^k}{(2k)!}$.
In view of (4) and (5) the relation (3) is proved. 
As a byproduct of the proof one obtains a simple expression for the coefficients of the series expansion (1):
$$
a_k=[x^{2k}]\frac{\sin nx}{\sin x}=[x^{2k}]\sum_{\ell=0}^{n-1}e^{i(n-1-2\ell)x}=\frac{(-1)^k}{(2k)!}\sum_{\ell=0}^{n-1}(n-1-2\ell)^{2k}.\tag6
$$
Particularly, $a_0=n$, $a_1=-\frac{n(n^2-1)}6$, and so on.
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