How prove this $\tan{x}>\frac{3x}{2+\sqrt{1-x^2}}$ let $0<x<1$,prove that
$$\tan{x}>\dfrac{3x}{2+\sqrt{1-x^2}}$$
This problem have nice solution?
my idea:
 let $$f(x)=\tan{x}-\dfrac{3x}{2+\sqrt{1-x^2}}=\tan{x}-3x\dfrac{2-\sqrt{1-x^2}}{3+x^2}$$
and  other idea:
let $x=\cos{t}$,then
$$\tan{(\cos{x})}>\dfrac{3\cos{t}}{2+\sin{t}}$$
other (2) idea:
$$\tan{x}>x+\dfrac{1}{3}x^3$$

why dont you go for the calculus approach. 
assume $ f(x) = tanx-3x/2+\sqrt{1+x^2} $. 
differentiate w.r.t to x 
we get $ f'(x) = sec^{2}x - 3/(2+\sqrt{1+x^2} + 3(x^2)/(2+\sqrt{1+x^2})^2 * 1/\sqrt{1+x^2} $ , 
under the given interval , $ f'(x) > 0 $ , implying it is an increasing function. 
   we know that $f'(x) > 0   <=>    f(x) > 0 .$. 
thus $ tanx > 3x/1+\sqrt{+x^2} $ 
