proof $|\sinh(x)|\leq|3x|$ for $ -\frac{1}{2}

Question

I am supposed to prove that $|\sinh(x)|\leq3|x|$ for $|x|<\frac{1}{2}$.

I know I am supposed to use $|e^x-1|\leq3|x|$ for $|x|<\frac{1}{2}$.

I am completely stuck, and I don't know how to approach this, so any help is greatly appreciated!

Answer 1

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le\frac{1}{1-x}} \tag 1$$

for $x<1$.

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Let $f(x) = 6x-e^x+e^{-x}$. Then, applying $(1)$ to $f(x)$, we find

$$\begin{align} 6x-e^x+e^{-x}&\ge 6x-\frac{1}{1-x}+(1-x)\\\\ &=\frac{x}{1-x}(4-5x)\\\\ &\ge 0 \end{align}$$

for $0\le x\le 4/5$.

Hence,

$$\bbox[5px,border:2px solid #C0A000]{|\sinh(x)|\le |3x|}$$

for $|x|\le 4/5$ as was to be shown!

Answer 2

Since $|\sinh{(-x)}|=|\sinh{x}|$, it's enough to prove that $0\leq\sinh{x}\leq3x$ for $0\leq x<\frac{1}{2}$ or $$0\leq e^x-e^{-x}\leq6x.$$ The left inequality is true because $e^x-e^{-x}=\frac{(e^x-1)(e^x+1)}{e^x}\geq\frac{(e^0-1)(e^x+1)}{e^x}=0.$

We'll prove the right inequality.

Let $f(x)=6x-e^x+e^{-x}$.

Hence, $$f'(x)=6-e^x-e^{-x}=-\frac{e^{2x}-6e^x+1}{e^x}=\frac{(3+2\sqrt2-e^x)(e^x-(3-2\sqrt2))}{e^x}\geq$$ $$\geq\frac{(3+2\sqrt2-e^{\frac{1}{2}})(e^0-(3-2\sqrt2))}{e^x}>0,$$ which says that $f$ is an increasing function on $[0,\frac{1}{2}]$.

Thus, $f(x)\geq f(0)=0$ and we are done!