How to prove the inequality $x^2 -\tan(x)\sin(x)<0$? Can somebody help me in proving $x^2 -\tan(x)\sin(x)<0$ in the interval $(0, \pi/2)$ by using first derivative concept?
It is easy to prove $x-\tan(x)<0$ or $\sin(x)-x<0$ in interval $(0, \pi/2)$ by using first derivative (by proving them monotonic). Your help will be appreciated.  
 A: Let $f(x)=x^2-\sin x\tan x$. Then $f'(x)=2x-\tan x\sec x-\sin x$. Now note that $\tan x\sec x+\sin x\ge 2\sqrt{\tan x\sec x\sin x}=2\tan x>2x.$
A: Due to the Cauchy-Schwarz inequality (CS) we have:
$$\sin(x)\tan(x)=\int_{0}^{x}\cos(t)\,dt \int_{0}^{x}\frac{dt}{\cos(t)^2}\stackrel{CS}{\geq}\left(\int_{0}^{x}\frac{dt}{\sqrt{\cos t}}\right)^2 $$
but over the interval $\left(0,\frac{\pi}{2}\right)$ we have $\cos t<1$, hence:
$$ \color{red}{\sin(x)\tan(x)}\geq\left(\int_{0}^{x}\frac{dt}{\sqrt{\cos t}}\right)^2\color{red}{>}\left(\int_{0}^{x}dt\right)^2 = \color{red}{x^2}$$
as wanted.

Since over the interval $\left(0,\frac{\pi}{2}\right)$ we also have $\cos t<1-\frac{4t^2}{\pi^2}$ by convexity, the same technique leads to the improved inequality:
$$ \forall x\in\left(0,\frac{\pi}{2}\right),\qquad \sin(x)\tan(x)>\frac{\pi^2}{4}\arcsin^2\left(\frac{\pi x}{2}\right).$$
A: The derivative of the function is
$f'(x)=2x-cos(x)tan(x)-\frac{tan(x)}{cos(x)}$
Rewrite to get
$f'(x)=2x-tan(x)(cos(x)+\frac{1}{cos(x)})$
Now, there is a famous inequality that you can use here.
$2\leq a+\frac{1}{a}$ for all positive real $a$
Since $cos(x)$ is positive over the interval $[0,\frac{\pi}{2})$
$2\leq cos(x)+\frac{1}{cos(x)}$
Therefore
$2(x-tan(x))\geq 2x-tan(x)(cos(x)+\frac{1}{cos(x)})$
as you have mentioned,
$x−tan(x)<0$
Hence, the derivative is always negative on the interval and knowing that the function starts from $0$, when $x=0$, it would always be less than $0$
