<s>[INST] 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:
$$\tan{x}>x+\dfrac{1}{3}x^3$$
 [/INST] \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. 
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