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:
$$\tan{x}>x+\dfrac{1}{3}x^3$$
 A: \We show equivalently that if $0\lt x\lt 1$, then
$$x\cot x\lt \frac{2+\sqrt{1-x^2}}{3}.\tag{1}$$
We use the fact that $x\cot x\lt 1-\frac{x^2}{3}$ in our interval. So we want to show that 
$$1-\frac{x^2}{3}\lt \frac{2+\sqrt{1-x^2}}{3}.\tag{2}$$
Multiply through by $3$, and rearrange. Inequality (2)  is equivalent to $1-x^2\lt \sqrt{1-x^2}$, which is clear. 
A: Your idea works equally well too.
We need to prove that $$ \cos{t} + \frac{\cos{t}^3}{3} > \frac{3\cos{t}}{2+\sin{t}}  $$
Since $x$ varies between $(0,1)$,  both  sin(t) and cos(t) vary from (0,1).
That reduces our inequality to:
$$ 1 + \frac{\cos^2{t}}{3} > \frac{3}{2+\sin{t}} $$
Expressing cos in terms of sin, and further simplifying we get:
$$ (2+\sin{t})^2(2-\sin{t}) > 3 $$
We substitute $y = \sin{t}$, where $y \in (0,1)$:
$$ -y^3 -2y^2 + 4y + 8 > 3 $$
Analyzing the derivative of the LHS, we can easily find that the minimum value of the polynomial in $(0,1)$ is $8$, and therefore the inequality is true in the given interval.
A: 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} $ 
