Prove that $\frac{\tan x}{x}>\frac{x}{\sin x}, x\in(0,\pi/2)$ 
Prove that $$\frac{\tan x}{x}>\frac{x}{\sin x},\;\;\; x\in(0,\pi/2).$$

My work
I formulated $$f(x)=\tan x \sin x - x^2$$ in hope that if $f'(x)>0$ i.e. monotonic then I can conclude for $x>0, f(x)>f(0)$ and hence, prove the statement.
However, I got $$f'(x)=\sin x + \sec x \tan x -2x, $$ where I am unable to conclude if $f'(x)>0.$
I also found $$f''(x)=\cos x + 2\sec^3x-\sec x-2,$$
$$f'''(x)=-\sin x (1-6\sec^4x+\sec^2x).$$
But I am not able conclude the sign of any of the higher derivatives either. Am I doing something wrong? Or is there some other way?

We can prove a stronger inequality 
$$\frac{\tan x}{x} > \left(\frac{x}{\sin x}\right)^2$$ for $x \in (0, \pi/2)$. Indeed the function 
$\tan x \sin^2 x - x^3$ has derivative $\tan^2 x + 2 \sin^2 x - 3 x^2$. Now, from this answer we see that $\frac{\tan x + 2 \sin x }{3} > x$ for $x \in (0, \frac{\pi}{2})$ so 
$\frac{\tan^2 x + 2 \sin^2 x}{3}> \left(\frac{ \tan x + 2 \sin x }{3}\right)^2 > x^2$ and  we are done. 
Note that the Taylor series of $1-\frac{x^3}{\tan x \sin^2 x}$ has all coefficients positive
$$1-\frac{x^3}{\tan x \sin^2 x}=\frac{x^4}{15} + \frac{4 x^6}{189} + \frac{x^8}{225}+\cdots$$
We will prove a weaker stamement than the above (which was only checked for some coefficients) by proving that the function 
$$f(x) = \frac{\sin x}{\cos^{\frac{1}{3}} x}$$ has the Taylor series at $0$ with all coefficients positive. For this, we notice that 
$$f^{(2)}(x) =\frac{4}{9} \frac{\sin^3 x}{\cos^{\frac{7}{3}}x} =\frac{4}{9} \sec^{\frac{4}{3}}x \cdot f^3(x)$$
Now, $\sec x$ at any positive power has all Taylor coefficients positive. This follows from the fact that $\sec x$ has a product expansion with factors $\frac{1}{1- a_k x^2}$ and the series $(1-t)^{-\alpha}$ has all Taylor coefficients positive. Now, the fact that $f$ has all coefficients positive follows by induction, by using the recurrence given by the differential equation. 
