Prove that $\tan x < \frac{4}{\pi}x,\forall x\in \left( 0;\frac{\pi}{4} \right)$ 
Prove that $$\tan x < \frac{4}{\pi}x,\forall x\in \left( 0;\frac{\pi}{4} \right)$$

I have known the solution that uses convex function. But I'd like another solution don't use it. :D
 A: Let $g$ be the function defined on the interval $[0,\pi/4]$ as 
$$g(x)=
\begin{cases}
\frac{\tan x}{x}&,x\ne 0\\\\
1&,x=0
\end{cases}
$$
Then, the derivative $g'$ of $g$ is given by
$$g'(x)=\frac{x\sec^2x-\tan x}{x^2}=\frac{x-\frac12\sin (2x)}{x^2\cos^2x}>0$$
for $x>0$ and $g'(0)=0$.  
Inasmuch as $g$ is increasing for $x\in[0,\pi/4]$ it attains, therefore, its maximum there at $x=\pi/4$.  Thus, 
$$g(x)<g(\pi/4)\implies \frac{\tan x}{x}<\frac{1}{\pi/4}\implies \tan x<\frac{4}{\pi}x$$
and we are done!
A: For $\lvert x \rvert <\pi/2$ the LHS is equal to $$\sum_{n=1}^\infty \frac{ B_{2n} (-4)^n (1-4^n) }{(2n)!} x^{2n-1},$$ so dividing by $x$ we get $$\sum_{n=1}^\infty \frac{ B_{2n} (-4)^n (1-4^n) }{(2n)!} x^{2n-2} < \frac{4}{\pi},$$ where the $B_{2n}$ are the Bernoulli numbers. Now, all the terms of the series are positive (
$1-4^n$ is always negative and exactly one of $B_{2n}$ and $(-4)^n$ is as well), so the LHS must be increasing and as a result we can just consider a narrow neighbourhood of $\pi/4$. The inequality follows from combining this with the fact that we know a priori that the LHS approaches $4/\pi$ as $x\to \pi/4$, and that the limiting value of the series is approached strictly from below (all the terms are positive).
A: If you study the function $g : x \mapsto \tan(x) - \frac{4}{\pi}x$, its derivative has a sole zero-point $x_0 = \arccos (\frac{\sqrt{\pi}}{2})$ thus it is decreasing in $(0,x_0)$ and increasing in $(x_0,\frac{\pi}{4})$. Since $g(0) = g(\frac{\pi}{4}) = 0$, $\forall x \in (0,\frac{\pi}{4}),\,\,  g(x) < 0$.
A: 
Here Slope of Line $\bf{OA}$ is $$\displaystyle \bf{m_{OA}} = \frac{1-0}{\frac{\pi}{4}-0} = \frac{4}{\pi}$$
And 
Here Slope of Line $\bf{OB}$ is $$\displaystyle \bf{m_{OB}} = \frac{\tan x-0}{x-0} = \frac{\tan x}{x}$$
so from Figure, We get $\displaystyle \bf{m_{OA}>m_{OB}}$
So we get $$\displaystyle \frac{4}{\pi}>\frac{\tan x}{x}\Rightarrow \tan x<\frac{4}{\pi}\cdot x \; \forall \; x\in \left(0,\frac{\pi}{4}\right)$$
