Prove that: $\forall x\in (0,\frac{ \pi}{4(n-1)})$ $\tan(nx)> n \tan(x)$ Let $n\in \mathbb{N} , n> 1$
Prove that : $\forall x\in (0,\frac{ \pi}{4(n-1)})$    $\tan(nx)> n \tan(x)$
I know: $f(x) = \tan x$  is convex function
$f(a x + b y) < a f(x) + b f(y),   a+b=1,  0< a, 0 < b,  x \neq y$
and what next?
 A: The interval for which the inequality is to hold seems to be not as large as possible, maybe since in the answer linked to by user lab bhattacharjee the method of induction was used to prove it.
So suppose instead that $nx$ is in the interval $(0,\pi/2)$ where the graph of tangent is strictly concave up. Because of this concavity we have
$$\tan kx-\tan (k-1)x > \tan (k-1) x - \tan(k-2)x,\tag{1}$$
for $k \ge 2.$ In particular each difference exceeds $\tan x-\tan 0 = \tan x,$ and then adding up the differences $\tan kx -\tan (k-1)x$ over $k=1 ... n$ the terms telescope to $\tan nx,$ which therefore exceeds $n \tan x$ because each of the $n$ terms summed exceeds it.
An added note: The method used here works just as well on any function $f$ which is say continuous on an interval $[0,a)$ and strictly concave up on $(0,a)$, with also $f(0)=0.$ It then allows, for $n>1$ and $x>0$ for which $nx<a,$ the conclusion that $f(nx)>nf(x).$
A: We have the well-known identity 
$$\tan(a+b) = \frac{\tan a+\tan b}{1-\tan a\tan b}. \tag{1}$$
If both $0<a,b<\frac\pi4$, then $0<\tan a<1$ and $0<\tan b<1$. This implies $0<1-\tan a\tan b<1$ and
$$\tan (a+b) > \tan a + \tan b. \tag{2}$$

Using the above inequality, we can show $$\tan nx > n\tan x$$ for $x<\frac\pi{4(n-1)}$ by induction. 
Sketch of the proof. To do this, we fix some $x$ from the given interval and prove (by induction on $k$) that $\tan kx > k\tan x$ for $k=2,3,\dots,n$. 
Notice that in the inductive step we are using the inequality $(2)$ for $a=(k-1)x$ and $b=x$, so the assumption that $a,b\in(0,\pi/4)$ is fulfilled.
Details of the induction.
$1^\circ$ Basis of the induction: $k=2$. Using $(2)$ for $a=x$ and $b=x$ we get
$$\tan(2x)>\tan x+\tan x.$$
Since $k=2\le n$, we have $n\ge 2$ and thus $a=b=x<\frac\pi{4(n-1)}<\frac\pi4$, so the assumptions of the inequality $(2)$ are fulfilled.
$2^\circ$ Inductive step. We assume that $k\le n$ and that
$$\tan (k-1)x>(k-1)\tan x.$$
If we apply $(2)$ to $a=(k-1)x$ and $b=x$, we get
$$\tan kx > \tan (k-1)x +\tan x > (k-1)\tan x+\tan x = k\tan x.$$
Again we should check that the assumptions $a,b<\frac\pi4$ are fulfilled, if we want to use $(2)$. It is clear that this is true for $b$. For $a$ we have
$$a=(k-1)x \le (n-1) x < \frac\pi4.$$
So $a,b\in(0,\pi/4)$ and we can indeed use the inequality $(2)$.
$\square$

As mentioned in another answer, this result can be in fact proved for a larger interval. If we want to obtain the same result using this approach, we could try to do it by showing that $(2)$ holds for $a,b>0$ such that $$a+b<\frac\pi2.$$
This can be done by observing that for such $a$ and $b$ we have $\tan b<\tan \left(\frac\pi2-a\right) = \frac1{\tan a}$, which means that $\tan a\tan b<1$ and $$0<1-\tan a\tan b<1.$$
So the validity of $(2)$ for $a$ and $b$ from this range can be shown using the same argument as in the first paragraph.
In fact, the same result can be shown for any convex function such that $f(0)=0$. See this answer for a proof of a non-strict inequality corresponding to $(2)$ in the case of concave function.
