Chứng minh rằng: 
a)   $\sin x>x-\frac{x^3}{6}, \forall x \in (0;\frac{\pi}{2})$
b)   $ \sin x+\tan x>2x, \forall x \in (0;\frac{\pi}{2})$
c)   $2\sin x+\tan x>3x, \forall x \in (0;\frac{\pi}{2})$
a)   Xét  $f(x)=\sin x-x+\frac{x^3}{6}, \forall x \in R$
$f'(x)=\cos x-1+-2\sin^2 \frac{x}{2}+\frac{x^2}{2}$
$=\frac{1}{2}[\frac{x^2}{2}-(\sin \frac{x}{2})^2]>0,  \forall x >0$  ( Vì $|\sin t|<t, \forall x >0$)
$\Rightarrow  f(x)>f(0)=0, \forall x >0$      $\Rightarrow \sin x>x-\frac{x^3}{6},  \forall x >0$

 b) Xét      $f(x)=\sin x+\tan x-2x,    x\in (0;\frac{\pi}{2})$
 $ f'(x)=\cos x+\frac{1}{\cos^2 x}-2>\cos^2 x+\frac{1}{\cos^2 x}-2$
            $ \geq (\cos x-\frac{1}{\cos x})^2\geq 0,  \forall  x\in (0;\frac{\pi}{2}) $
 $\Rightarrow f(x)>f(0)=0, \forall  x\in (0;\frac{\pi}{2})  \Rightarrow \sin x+\tan x>2x, \forall  x\in (0;\frac{\pi}{2})$

 c) Xét    $f(x)=2\sin x+\tan x-3x, \forall  x\in (0;\frac{\pi}{2})$
 $f'(x)=2\cos x+\frac{1}{\cos^2 x}-3=\cos x+\cos x+\frac{1}{\cos^2 x}-3$
             $>3\sqrt[3]{\cos x.\cos x.\frac{1}{\cos^2 x}}-3=0,  \forall  x\in (0;\frac{\pi}{2})$
 $\Rightarrow f(x)>f(0)=0,  \forall  x\in (0;\frac{\pi}{2})$
 $\Rightarrow 2\sin x+\tan x>3x,  \forall  x\in (0;\frac{\pi}{2})$

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