Using Mean Value Theorem, Prove that ${\tan x\over x}>{x\over\sin x}$ Prove using Mean Value Theorem,

$${\tan x\over x}>{x\over\sin x} \space\forall \space x \space\in(0, \pi/2) $$

Attempt::
$f(x) = x-\sin x$
$f^\prime(x) = 1-\cos x > 0 $
Hence $x-\sin x>0, {x\over\sin x}>1$
Similarly, I got ${\tan x\over x}>1$
But how do I compare them and get the required inequality?
 A: We need to prove that $f(x)>0$, where $f(x)=\frac{\sin{x}}{\sqrt{\cos{x}}}-x$.
But $f'(x)=\frac{\cos x\sqrt{\cos x}+\frac{\sin^2x}{2\sqrt{\cos x}}}{\cos{x}}-1=\frac{(\sqrt{\cos^3{x}}-1)^2+\cos^2x(1-\cos{x})}{2\sqrt{\cos^3x}}>0$.
Thus, $f(x)>f(0)=0$.
Done!
A: For each $x \in (0,2\pi),$ the MVT(mean value theorem) makes sure that there exists $c_x$ with $0 < c_x < x$ such that
$$\frac{\sin x}{x} = \cos c_x < 1.$$
With $f(x) = \sin x$ we have $\sin x = \sin x - \sin 0 = f'(c_x)(x-0)= (\cos c_x)x$ 
where $0 < c_x < x$.
Now for $\displaystyle \frac{\tan x}{x}>1$ 
$f(x) = \tan x$ then $f'(x) = \sec^2(x)$.  So $\sec^2(x)$ = $\displaystyle \frac{\tan x}{x}$
For each $x \in (0,2\pi)$ $\sec^2(x) >1$. So $\displaystyle \frac{\tan x}{x}$ $>1$.
A: Suppose we have a function $f(t)=\sqrt{\sin{t}\tan{t}}$. Now, applying the mean value theorem in $(0, \frac{\pi}{2})$, we have $$\frac{f(x)-f(0)}{x-0}=f^\prime(c)$$ Or $$\frac{f(x)}{x}=f^\prime(c)\quad (where\quad 0<c<x<\frac{\pi}{2})$$ Or $$\frac{\sqrt{\sin{x}\tan{x}}}{x}=f^\prime(c)\tag{1}$$ Now we proceed to derivation of $f(t)$: $$f^\prime(t) = \frac{\sin{t}\sec^2{t}+\tan{t}\cos{t}}{2\sqrt{\sin{t}\tan{t}}}$$ $$=\frac{\tan{t}(\sec{t}+\cos{t})}{2\sqrt{\sin{t}\tan{t}}}$$ $$=\frac{\tan{t}(\sec{t}+\cos{t})\sqrt{\cos{t}}}{2{\sin{t}}}$$ $$=\frac{\sec{t}+\cos{t}}{2\sqrt{\cos{t}}}$$ $$=\frac{\sec{t}\sqrt{\sec{t}}+\sqrt{\cos{t}}}{2}$$ Now, $$f^{\prime\prime}(t)=\frac{3\sec{t}\tan{t}-\sin{t}}{2{\sqrt{\cos{t}}}}$$ Since $f^{\prime\prime}(t) = 0$ at $ t = 0$ we get that $f^\prime(t)$ is minimum at $t = 0$. As $f^\prime(0) = 1$, we conclude that for $c\in(0, \frac{\pi}{2})$, $f^\prime(c)>1$. Therefore, from (1) we have, $$\frac{\sqrt{\sin{x}\tan{x}}}{x}>1\quad x\in(0, \frac{\pi}{2})$$ or $$\frac{\sin{x}\tan{x}}{x^2}>1$$ or $$\frac{\tan{x}}{x}>\frac{x}{\sin{x}}\quad
 if \quad 0<x<\frac{\pi}{2}$$
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