Proving $\frac{\sqrt3\cos x-\sin x}{\sin 3x}> \frac{\sqrt3}{3x}-\frac13$ for small $x>0$ 
Prove that$$\frac{\sqrt3\cos x-\sin x}{\sin 3x}> \frac{\sqrt3}{3x}-\frac13$$
  for small $x$ near $0$.

From Taylor expansion I can see that $$\frac{\cos x}{\sin3x}>\frac13x,\quad \frac{\sin x}{\sin3x}>\frac13,$$ but combining these two does not give actually what I want. I kindly thank you anyone that can provide at least an idea. I know this is true graphically but I want to see rigorously. 
 A: Using the identity from @labbhattacharjee, for $0 < x < \dfrac{π}{3}$ there is\begin{align*}
0 < \frac{\sin 3x}{\sqrt{3}\cos x - \sin x} &= \frac{1}{2} \frac{\sin\left( 3\left( \dfrac{π}{3} - x \right) \right)}{\sin\left( \dfrac{π}{3} - x \right)} = \frac{1}{2} \left( 3 - 4\sin^2\left( \dfrac{π}{3} - x \right) \right)\\
&= \frac{1}{2} \left( 1 + 2\cos\left( \dfrac{2π}{3} - 2x \right) \right) = \frac{1}{2} (1 - \cos 2x + \sqrt{3} \sin 2x)\\
&= \sin^2 x + \sqrt{3} \sin x \cos x = \sin x (\sin x + \sqrt{3} \cos x)\\
&< x(x + \sqrt{3}) = x \cdot \frac{3 - x^2}{\sqrt{3} - x} < \frac{3x}{\sqrt{3} - x},
\end{align*}
thus$$
\frac{\sqrt{3}\cos x - \sin x}{\sin 3x} > \frac{\sqrt{3} - x}{3x} = \frac{1}{\sqrt{3} x} - \frac{1}{3}.
$$
A: COMMENT.- 
$$\frac{\sqrt3\cos x-\sin x}{\sin 3x}> \frac{\sqrt3}{3x}-\frac13\iff\frac{3x}{\sin 3x}\cdot\frac{\sqrt3\cos x-\sin x}{\sqrt3-x}\gt1$$ 
Both factors in the LHS tend to $1$  and for small $x\gt0$ the first factor is greater than $1$ and the second factor is less than $1$ but it can be proved that for small $x\gt 0$ 
$$\frac{3x}{\sin 3x}\gt\frac{\sqrt3\cos x-\sin x}{\sqrt3-x}$$ 
(note that the denominator $\sqrt3-x$ is quite greater than $\sin 3x$ so you look at the numerators)
