Prove some inequalities using the mean value theorem

Prove the following inequalities using the mean value theorem:

$| \sin x - \sin y| \leq |x - y|$ .
$ny^{n-1} (x-y) \leq x^n - y^n \leq nx^{n-1} (x-y) \qquad \text{if } 0 < y \leq x$ , for $n = 1, 2, 3, \ldots$ .

Proof. Define $f(t) = \sin t$ and $g(t) = t$ . Then $f$ and $g$ are continuous and differentiable everywhere so we may apply the mean value theorem. We obtain
$\begin{align*} &&f'(c) (g(x) - g(y)) &= g'(c) (f(x) - f(y)) & (\text{for some } c \in (x,y))\\ \implies && (\cos c) (x-y) &= \sin x - \sin y \\ \implies && | \cos c | | x-y| &= | \sin x - \sin y| \\ \implies && |x-y| &\geq | \sin x - \sin y|. \end{align*}$

The final step follows since $| \cos c | \leq 1$ for all $c. \qquad \blacksquare$
Proof. Let $f(t) = t^n$ , $g(t) = t$ , then $f'(t) = nt^{n-1}$ and $g'(t) = 1$ . So, by the mean-value theorem we have there exists a $c \in (x,y)$ such that,
$\begin{align*} && f'(c) (g(x) - g(y)) &= g'(c)(f(x) - f(y)) \\ \implies && nc^{n-1}(x-y) &= x^n - y^n. \end{align*}$

But, since $x^{n-1}$ is an increasing function on the positive real axis, and we have $0 < y \leq c \leq x$ we know

$y^{n-1} \leq c^{n-1} \leq x^{n-1}.$

Further, since $(x-y) \geq 0$ and $n$ is positive we can multiply all of the terms in the equality by $n (x-y)$ without reversing inequalities to obtain,

$ny^{n-1}(x-y) \leq nc^{n-1} (x-y) \leq nx^{n-1} (x-y).$

Therefore, substituting $nc^{n-1} (x-y) = x^n - y^n$ from above we obtain the requested inequality:

$ny^{n-1} (x-y) \leq x^n - y^n \leq nx^{n-1} (x-y). \qquad \blacksquare$

Stumbling Robot