<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  $\frac{x^2n(n^2-1)}{3}\geqslant \cos{\left(a+\frac{(n-1)x}{2}\right)}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\overset{(4,x\to 2x)}{\iff}\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}} \square$


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*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] 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.
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