Answer
$sin~x \lt x~~$ if $~~0 \lt x \lt 2\pi$
Work Step by Step
Let $f(x) = sin~x-x$
Then $f'(x) = cos~x-1$
Note that $f'(x) \lt 0$ on the interval $(0,2\pi)$
1. $f$ is continuous on the interval $[0,2\pi]$
2. $f$ is differentiable on the interval $(0,2\pi)$
$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[0,2\pi]$
Choose any number $b$ in the interval $(0,2\pi)$
According to the Mean Value Theorem Theorem, there is a number $c$ in the interval $(0,b)$ such that $f'(c) = \frac{f(b)-f(0)}{b-0}$
Then:
$f'(c) = \frac{f(b)-f(0)}{b-0} \lt 0$
$f(b)-f(0) \lt 0$
$(sin~b-b)-(sin~0-0) \lt 0$
$(sin~b-b)-(0) \lt 0$
$sin~b-b \lt 0$
$sin~b \lt b$
Therefore, since the number $b$ was chosen arbitrarily from the interval $(0,2\pi)$, $~~sin~x \lt x~~$ if $~~0 \lt x \lt 2\pi$
