Deriving the inequality $|\sinh{x}| \le 3|x|, \quad |x| < 1/2$

Question

I will like to derive $$|\sinh{x}| \le 3|x|, \quad |x| < 1/2$$

I got this the idea of using this

$$|\exp{x}-1| \le 3|x|, \quad x < 1/2$$

which I found in the appendix. (I admit, that I was looking for something which looked like it).

Q1: The first inequality have the condition $|x| < 1/2$, but the second have $x < 1/2$. I guess that's OK.

Q2: I'm actually not sure if this is the right route I'm taking.

Q3: There is one hint: "use the addition formula". I'm not sure which formula? I think it's this: $$ \sinh (x+y)=\cosh x \sinh y+\sinh x \cosh y $$

I'm not sure how I will bring this in the play.

NB: Things which are not allowed: - mean-value theorem

The rule is that I have to use as little theory as possible.

Answer 1

How about using the mean-value theorem? There exists a point $\xi$ in between $x$ and $-x$ satisfying $$e^x - e^{-x} = e^{\xi}(x - (-x))$$ so that $$\frac{e^x - e^{-x}}{2} = e^\xi x.$$ Take the absolute value to find $$|\sinh x| \le e^\xi |x|.$$ As long as $|x| \le \log 3$ (which is in fact larger than $1$) you get $|\sinh x| \le 3|x|$.

Answer 2

A basic inequality is: \begin{equation} \tag{1}\label{3545836:eq:1} e^x \geqslant 1 + x, \end{equation} which holds for all real $x.$

This has been shown in several previous threads, such as: Simplest or nicest proof that $1+x \le e^x$.

For what it's worth, my favourite proof of \eqref{3545836:eq:1} is along similar lines to my previous answer to the present question: use the integral definition of the logarithm to prove \begin{equation} \tag{2}\label{3545836:eq:2} \frac{t- 1}t \leqslant \log t \leqslant t - 1 \quad (t > 0), \end{equation} distinguishing the cases $t \geqslant 1$ and $t \leqslant 1$ (using both inequalities in \eqref{3545836:eq:2}, we can deduce the case $t \leqslant 1$ from the case $t \geqslant 1,$ although this isn't necessary), and then put $t = e^x$ in the second inequality. $\square$

Replacing $x$ by $-2x$ in \eqref{3545836:eq:1}, we get: $$ 1 - e^{-2x} \leqslant 2x $$ for all real $x,$ and therefore: $$ \sinh x = \frac{e^x - e^{-x}}2 = e^x\frac{1 - e^{-2x}}2 \leqslant xe^x \leqslant ex \leqslant 3x \quad (0 \leqslant x \leqslant 1). $$

Answer 3

The function $x\to\dfrac{\sinh x}{x}$ is even and increasing for $x\ge0$, since it's a sum of increasing functions (write de Taylor series).

So all you have to prove is $\sinh (\dfrac12)<\dfrac32$.

But, for $0< x<1$,

$$\sinh x<x+x^3+x^5+\dots=\frac{x}{1-x^2}$$

Now let $x=\dfrac12$ in the inequality above.

Answer 4

As the functions $\sinh x$ and $3x$ are odd, it is enough to prove it for $0\le x<\frac12$.

Now, simply use that $\sinh x$ is a convex function on $[0,+\infty)$. This implies that, on this interval, the graph of the function is below the chord determined by the origin and the point with abscissa $\frac12$,. Explicitly, if $\;0\le x\le\tfrac12$: $$0\le \sinh x \le \frac{ \sinh\frac12}{\frac12}x=2\sinh\tfrac12 x<\mathrm e^{\tfrac12}x <3x.$$

Answer 5

I doubt if this method is allowed by the question, but it is appealingly geometrical. We prove: $$ x < \sinh x < 2x \quad (0 < x \leqslant \log2). $$

Proof. Write $u = e^x,$ so that $1 < u \leqslant 2,$ and $$ x = \log u = \int_1^u\frac{dt}t. $$ Thus $x$ is the area of the region $R$ bounded by the graph of $\frac1t,$ the $t$-axis, and the vertical lines $t = 1, t = u$.

This region properly contains the rectangle whose vertices are $$ (1, 0), \left(1, \frac12\right), (u, 0), \left(u, \frac12\right), $$ and whose area is $$ \frac{u-1}2, $$ therefore $$ u - 1 < 2x. $$

The function $\frac1t$ is convex, so the region $R$ is properly contained in the trapezium $T$ whose vertices are $$ (1, 0), (1, 1), (u, 0), \left(u, \frac1u\right), $$ and whose area is $$ \frac{u-1}2\left(1 + \frac1u\right) = \frac{u^2 - 1}{2u} = \frac{e^x - e^{-x}}2 = \sinh x. $$ But $$ 1 + \frac1u < 2 $$ (more geometrically: the trapezium $T$ is properly contained in the larger rectangle whose vertices are $(1, 0), (1, 1), (u, 0), (u, 1),$ and whose area is $u - 1$), therefore $$ x < \sinh x < u - 1 < 2x. $$