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 $\pi$).



 A: 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.
A: Remark: My previous solution is ugly. I give another solution.
Clearly, we only need to prove the case when $x > 0$.
For $n=1$, clearly the inequality is true.
For $n=2$, the inequality is equivalent to $2\cos x - 2 + x^2\ge 0$ which is true.
For $n\ge 3$ and $x\in [\frac{\pi}{2}, \infty)$:
It is easy to prove that
$-n \le \frac{\sin nx}{\sin x} \le n$ (by math induction) and $-n \ge n - \frac{n(n^2-1)x^2}{6}$. The inequality is true.
For $n \ge 3$ and $x\in (0, \frac{\pi}{2})$: Let $f(n) \triangleq \frac{\sin nx}{\sin x} - n + \frac{n(n^2-1)x^2}{6}$.
We have 
\begin{align}
f(n+2) - 2f(n+1) + f(n) &= \frac{-4(\sin\frac{x}{2})^2\sin (n+1) x}{\sin x} + (n+1)x^2\\
&\ge -4(n+1)(\sin\frac{x}{2})^2 + (n+1)x^2\\
& \ge 0
\end{align} 
where we have used 
$\sin (n+1) x \le (n+1)\sin x$.
Thus, we have $f(n+2) - f(n+1) \ge f(n+1) - f(n)$ for all $n\ge 3$.
Thus, we have 
\begin{align}
f(n+1) - f(n) &\ge f(4) - f(3) \\
&= 8(\cos x)^3- 4(\cos x)^2+6x^2-4\cos x\\
&= 2(4\cos x + 3)(1-\cos x)^2 + 6x^2 - 6(\sin x)^2\\
& \ge 0.
\end{align}
Thus, we have $f(n) \ge f(3) = -4(\sin x)^2+4x^2\ge 0$ for all $n\ge 3$.
We are done.
A: Let $f(n,x)=n-\frac{(n-1)n(n+1)}{6}x^2$ the RHS of the required inequality. Wlog we can ssume $n \ge 3$ as for $n=1,2$ the inequality is trivial ($n=2$ reduces to $2\sin^2(\frac{x}{2}) \le \frac{x^2}{2}$ which follows from $|\sin x| \le |x|$).
Using the well known inequality $|\frac{\sin nx}{\sin x}| \le n$, the required inequality is non-trivial and needs to be proven only for $n-\frac{(n-1)n(n+1)}{6}x^2 \ge -n$ or $x^2 \le \frac{12}{n^2-1}$ which means $x^2 \le \frac{3}{2}$  when $n \ge 3$ so definitely $|x| < \frac{\pi}{2}$ so $\cos x >0$.
Note that $\frac{\sin(n+1)x}{\sin x}-f(n+1,x)=(\frac{\sin(n)x}{\sin x}-f(n,x))\cos x-2\sin^2(\frac{x}{2})f(n,x)-2\sin^2(\frac{nx}{2})+\frac{n(n+1)}{2}x^2$
So assuming by induction $\frac{\sin(n)x}{\sin x}-f(n,x) \ge 0$ (starting at $n=2$ as $2x^2 \ge 3$ hence $\cos x>0$ when $n+1 \ge 3$ while the inequality is trivial otherwise as seen above) and noticing that $-2\sin^2(\frac{nx}{2}) \ge -\frac{n^2x^2}{2}$. while if $f(n,x) \ge 0$ then $f(n,x) \le n$ so in any case $-2\sin^2(\frac{x}{2})f(n,x) \ge -\frac{n}{2}x^2$, we get:
$\frac{\sin(n+1)x}{\sin x}-f(n+1,x) \ge 0$ and we are done!
A: Alternative solution:
Remark: $n - \frac{n(n^2-1)x^2}{6}$ is the second order Taylor approximation of $f(x) = \frac{\sin nx}{\sin x}$ around $x = 0$.
First, we give the following result. The proof is given later.
Fact 1: Let $n\ge 3$ be a positive integer and $x\in (0, \frac{\pi}{2})$. Then 
$\frac{\sin nx}{\sin x} \ge n - \frac{n(n^2-1)x^2}{6}$.
The remaining cases are easy to prove:
For $n=1$, clearly the inequality is true.
For $n=2$, the inequality is equivalent to $2\cos x - 2 + x^2\ge 0$ which is true.
For $n\ge 3$ and $x\in [\frac{\pi}{2}, \infty)$, 
we have $-n \ge n - \frac{n(n^2-1)x^2}{6}$.
It is easy to prove that $-n \le \frac{\sin nx}{\sin x} \le n$ for $x\in \mathbb{R}$ (by math induction).
The inequality is true.
$\phantom{2}$
Proof of Fact 1: The inequality is written as
$$\frac{\sin n x}{nx} \ge \frac{\sin x}{x} - \frac{n^2-1}{6}x^2\cdot \frac{\sin x}{x}.$$
To proceed, we need the following results (Facts 2 through 3). Their proof is not hard and hence omitted.
Fact 2: $\frac{\sin y}{y} \ge \frac{-7y^2+60}{3y^2 + 60}$ for $y\in \mathbb{R}$.
(Pade $(2,2)$ approximation)
Fact 3: $\frac{\sin y}{y} \le \frac{6}{6+y^2}$ for $y \in (0, \frac{\pi}{2})$.
(Pade $(0,2)$ approximation)
By using Facts 2 and 3, it suffices to prove that
$$\frac{-7n^2x^2+60}{3n^2x^2 + 60} 
\ge \frac{6}{6+x^2} - \frac{n^2-1}{6}x^2\cdot \frac{-7x^2+60}{3x^2 + 60}$$
or (after clearing the denominators)
$$(-7x^4+18x^2+360)n^4+(7x^4-200x^2-1200)n^2+140x^2 \ge 0.$$
Since $-7x^4+18x^2+360 > 0$, we have
\begin{align}
&(-7x^4+18x^2+360)n^4+(7x^4-200x^2-1200)n^2+140x^2\\
\ge \ & (-7x^4+18x^2+360)n^2\cdot 3^2+(7x^4-200x^2-1200)n^2+140x^2\\
= \ & (-56x^4-38x^2+2040)n^2+140x^2\\
\ge \ & 0
\end{align}
where we have used the fact that $-56x^4-38x^2+2040 > 0$. We are done.
