How to show that $\tanh(x) := \frac{e^{2x}-1}{e^{2x}+1}$ is strictly monotically increasing $\tanh(x) := \frac{e^{2x}-1}{e^{2x}+1}$
Wether a function increases or descreases can be found by taking a look at the derivative. If the derivative is $>0 \forall x \in \mathbb{R}$ then it is monotonically increasing.
$\tanh'(x) := \frac{4e^{2x}}{(e^{2x}+1)^2}$
Im not sure if this part is correct at all, or if yes, wether it can be done in a simpler way:
$$\frac{4e^{2x}}{(e^{2x}+1)^2} \le^! 0 \Leftrightarrow 4e^{2x} \le 0 \Leftrightarrow 1 \le 0 $$
Because $1 > 0$, $\frac{4e^{2x}}{(e^{2x}+1)^2} > 0$ $\forall x \in \mathbb{R}$,  $\tanh(x)$ is monotonically increasing.
 A: Hint:
$$\tanh(x)=\frac{e^{2x}+1-2}{e^{2x}+1}=1-\frac{2}{e^{2x}+1}$$
Note that we are subtracting a number that gets smaller as we increase $x$
A: Because it is the composition of 2 strictly increasing functions $x \to y=\dfrac{x-1}{x+1} $ and $x \to e^{2x}$.
A: Alternatively:
$$y'=\left(\tanh{x}\right)'=\frac{1}{\cosh^2{x}}>0, x\in R \Rightarrow y \uparrow$$
A: There is a simpler way than going via your contradiction. Just note that $e^u$ is always positive for real $u$, and $4$ is always positive, and $1$ is always positive, and sums, products and fractions of positive numbers are always positive. Therefore $\frac{4e^{2x}}{(e^{2x}+1)^2}>0$ for all $x$.
(You would have to show all of this to prove that your inequality manipulations are valid anyways, it is only because these are all positive that you can multiply / divide them away without changing the direction of the inequality.)
A: Let $0<x<y$. Limiting to $\sinh x$;
Then
$$\implies e^x(1-e^{-x-y})<e^y(1-e^{-x-y})$$
Which means we only need to show for arbitrary $x, y$ that $e^x < e^y$. If I were begin totally robust I would have started with the latter inequality to propose the first. however:


*

*If $(1-e^{-x-y}) = 0$ then $e^x(1-e^{-x-y})=e^y(1-e^{-x-y})$.

*If $(1-e^{-x-y}) < 0$ and $e^x < e^y$  then $e^x(1-e^{-x-y}) > e^y(1-e^{-x-y})$.


From here you should conclude that $\sinh x$ is strictly monotone, for $\tanh x$ you can use the simple relationship that $\tanh^2 x+ \operatorname{sech}^2 x=1$.
