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Prove some inequalities using the mean value theorem

Prove the following inequalities using the mean value theorem:

  1. | \sin x - \sin y| \leq |x - y|.
  2. 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.

  1. 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

  2. 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:

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


    • William says:

      I’m pretty sure you’re right, but for the inequality y^{n-1} <= c^{n-1} \<= x^{n-1}, I think we have to have <= signs because n could equal one (even though we know n is strctly between x and y), so we still end up with less than or equal to signs in the solution

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