Show the inequalities $|\sinh x| \leq 3|x|$ and $| \cosh x - 1 | \leq 3|x|$ for $|x| \lt \frac{1}{2}$

Question

I am working through a real analysis book and one of the exercises want me to show that these inequalities exists.

$|\sinh x| \leq 3|x|$ and $| \cosh x - 1 | \leq 3|x|$ for $|x| \lt \frac{1}{2}$

I have been banging my head against this for almost an hour, trying to algebraically manipulate the terms to show this, but I have had no luck. Can anyone help me or give me tips on how to solve this?

Answer 1

Use power series:\begin{align}\lvert\sinh x\rvert&=\left\lvert x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots\right\rvert\\&\leqslant\lvert x\rvert+\frac{\lvert x\rvert^3}{3!}+\frac{\lvert x\rvert^5}{5!}+\cdots\\&=\lvert x\rvert\left(1+\frac{\lvert x\rvert^2}{3!}+\frac{\lvert x\rvert^4}{5!}+\cdots\right)\\&\leqslant\lvert x\rvert\left(1+\left(\frac12\right)^2+\left(\frac12\right)^4+\cdots\right)\\&=\frac43\lvert x\rvert.\end{align}

Answer 2

For $|x|<\frac{1}{2}$ we have, by the Mean Value Theorem: there is $t$ such that $|t|<\frac{1}{2}$ and

$| \sinh x|= |\sinh x - \sinh 0|=|x| \cosh t.$

Can you proceed ?