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?
Best Answer
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!