Solving trigonometric inequality with elementary means In a set of olympiad problems, the last,hardest one was:

We know that $\sin(x)<x<\tan(x)$ for every real $x\in (0,\pi/2)$. What is the maximum integer value of $n$ such that holds for every $x\in(0,\pi/2)$
  $$\sin(2x)+\tan(2x)>nx$$

By taylor expansion and term by term sum the answer is immediatly $4$.
But what was the intended, elementary way of solving it(using the information they gave)? 
 A: I'm assuming that the maximal (integer) $n$ such that
$$\sin (2x) + \tan (2x) > nx$$
holds for $0 < x < \frac{\pi}{4}$ is sought, and not for $\frac{\pi}{2}$. Otherwise no such $n$ exists.
So looking at
$$\frac{\sin (2x)}{2x} + \frac{\tan (2x)}{2x} = \frac{\sin (2x)}{2x}\left(1 + \frac{1}{\cos(2x)}\right) < 1 + \frac{1}{\cos (2x)},$$
letting $x \searrow 0$ immediately shows $n \leqslant 4$. On the other hand,
$$\begin{align}
\frac{\sin (2x)}{2x} + \frac{\tan (2x)}{2x} &= \frac{\tan (2x)}{2x}\left(\cos (2x) + 1\right)\\
&= \frac{\tan (2x)}{2x}2\cos^2 x\\
&= \frac{\cos^2 x}{x}\cdot \frac{2\tan x}{1 - \tan^2 x}\\
&= 2\frac{\cos^2 x}{1-\tan^2 x}\cdot \frac{\tan x}{x}\\
&> 2 \frac{\cos^2 x}{1 - \tan^2 x}.
\end{align}$$
But we know that $0 < \sin x < \tan x$ for $0 < x < \pi/2$, so we have
$$\sin^2 x < \tan^2 x \iff 1 - \cos^2x < \tan^2 x \iff 1-\tan^2 x < \cos^2 x,$$
and for $0 < x < \pi/4$, we have $0 < 1 - \tan^2 x$, so
$$\frac{\cos^2 x}{1-\tan^2 x} > 1,$$
whence, for $0 < x < \pi/4$, we have
$$\frac{\sin (2x)}{2x} + \frac{\tan (2x)}{2x} > 2,$$
or
$$\sin (2x) + \tan (2x) > 4x.$$
A: By the trigonometric identities, We know that
$$sin(2x)=\dfrac{2tanx}{1+tan^2x}, \quad tan(2x)=\dfrac{2tanx}{1-tan^2x}$$
then, 
$$sin(2x)+tan(2x)=\dfrac{2tanx}{1+tan^2x}+\dfrac{2tanx}{1-tan^2x}$$
$$=\dfrac{4tanx}{1-tan^4x} \gt nx$$
Because $x \neq 0,$
$$\dfrac{4tanx}{x} \gt n(1-tan^4x)$$
From the problem, $x \lt tanx$, so
$$\dfrac {4tanx}{x} \gt 4$$
and since $tan^4x \gt 0 \quad x \in (0, \pi/2)$,
$$n \gt n(1-tan^4x)$$
Therefore, $4 \ge n$
