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I need help understanding this proof:

Prove that $\cosh$ is strictly increasing on $[0,\infty)$.

From $(\cosh x)^2=1+(\sinh x)^2\geq 1$ it follows that $R(\cosh)\subseteq [1,\infty)$. From that for $x,y>0$ we have: $\cosh (x+y)=\cosh x \cdot \cosh y+\sinh x\cdot \sinh y> \cosh x\cdot \cosh y> \cosh x$ so $\cosh$ is strictly increasing on $[0,\infty)$.

$1)$ Why is $R(\cosh)\subseteq[1,\infty)$? If $(\cosh x)^2\ge 1$ doesn't that mean $R(\cosh x)=(-\infty, -1]\cup [1,\infty)$? Then I would say $[1,\infty)\subseteq R(\cosh x)$ and not the other way around?

$2)$ I understand that $\cosh (x+y)=\cosh x \cdot \cosh y+\sinh x\cdot \sinh y> \cosh x\cdot \cosh y> \cosh x$ but I don't see how that proves $\cosh x$ is strictly increasing on $[0, \infty)$... If we want to prove that a function is strictly increasing, then I thought we should take $a, b \in D(f)$ and prove that if $a

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    For your first point, this depends on what else you know about $\cosh$. It is true that $f(z)^2≥1$ does not imply $f(z)\in[1,\infty)$. But you may know that $\cosh$ already is positive. For the second point let $a$b-a$ is positive and $\cosh(b)=\cosh(a+(b-a)) >\cosh(a)$ follows from the statement shown. – 2017-02-04

2 Answers 2

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In $1)$, the knowledge that $\cosh x \geqslant 0$ for all $x\in \mathbb{R}$ (or only for $x \geqslant 0$) is used. With that knowledge, from $(\cosh x)^2 \geqslant 1$ one can infer $\cos x \geqslant 1$, i.e. $R(\cosh) \subseteq [1,\infty)$. Indeed $R(\cosh) = [1,\infty)$, but showing equality requires a further step. For example with $f(x) = \sqrt{1 + \lfloor x^2\rfloor}$ we also can deduce $R(f) \subseteq [1,\infty)$ from the fact that $f(x) \geqslant 0$ for all $x$ and $f(x)^2 = 1 + \lfloor x^2\rfloor \geqslant 1$, but here $R(f) \subsetneq [1,\infty)$.

Regarding point $2)$, for $0 \leqslant a < b$, take $x = a$ and $y = b-a$, then the argument yields

$$\cosh b = \cosh (x+y) > \cosh x = \cosh a.$$

In the argument, the step $x > 0 \implies \cosh x > 1$ is however missing.

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Here is an alternative proof: Set $a=\exp x_1$ and $b=\exp x_2$ where $x_1,x_2 \in [0,\infty)$ and $x_1