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?

I believe the simplest proof is through the Cauchy-Schwarz inequality:
$$\tan(x)\sin(x)=\int_{0}^{x}\frac{d\theta}{\cos^2\theta}\int_{0}^{x}\cos(\theta)\,d\theta\geq\left(\int_{0}^{x}\frac{d\theta}{\sqrt{\cos\theta}}\right)^2\geq\left(\int_{0}^{x}d\theta\right)^2=x^2. $$
In a similar fashion, for any $x\in\left(0,\frac{\pi}{2}\right)$ we have  $\frac{\tan x }{x}\geq\left(\frac{x}{\sin x}\right)^2$ by Holder's inequality.
