Prove that $f(x)\leq\cosh(x)$ $\forall$ $ x\in\mathbb{R^+} $ 

Let $ f:\mathbb{R^+}\to(1,+\infty) $ be continuously differentiable function such that 
     $ f^{2}(x) -(f^{'})^{2}(x)\geq 1$ $\forall$ $ x\in\mathbb{R^+} $ and $ f(0) =1$.
Prove that  $f(x)\leq\cosh(x)$ $\forall$ $ x\in\mathbb{R^+} $


My proof :
Put $f(x)=\cosh(g(x))$ with $g(0)=0$ we get :
$$\cosh^2(g(x))-(g'(x))^2\sinh^2(g(x))\geq 1$$
Or :
$$-(g'(x))^2\sinh^2(g(x))\geq 1-\cosh^2(g(x))$$
Or:
$$-(g'(x))^2\sinh^2(g(x))\geq -\sinh^2(g(x))$$
Or:
$$(g'(x))^2\leq 1$$
Integrating we get :
$$|g(x)|\leq |x|$$
So we get the desired result since $\cosh(x)$ is increasing $\forall$ $ x\in\mathbb{R^+} $
My question
I'm really curious to see an alternative proof so : Have you another proof ?
Thanks a lot for all your contributions
 A: $\DeclareMathOperator{\acosh}{acosh}$
I think your solution is fine. $g(x) = \acosh(f(x))$ is continuous for $x \ge 0$, and differentiable and strictly positive for $x > 0$.  You derived that
$$
 \sinh^2 g(x) (g'(x)^2 - 1) \le 0
$$
so that we can conclude
$$
 g'(x)^2 \le 1 \implies g'(x) \le 1 
$$
for $x > 0$, and $g(x) \le x$ follows.
An alternative approach is to write
$$
 f'(x) \le |f'(x)| \le \sqrt{f^2(x) - 1}
$$
so that for $0 < x < y$
$$
 \int_x^y \frac{f'(t)}{\sqrt{f^2(t) - 1}} \, dt \le \int_x^y 1 \, dt \\
\implies \acosh(f(y)) - \acosh(f(x)) \le y - x \, .
$$
Letting $x \to 0$ we get
$$
\acosh(f(y)) \le y \implies f(y) \le \cosh(y) \, .
$$

If suffices to assume that $f:[0, \infty) \to \Bbb R$ is continuous with $f(0) = 1$, and that $f$ is differentiable on $(0, \infty)$ with $f(x)^2 - f'(x)^2 \ge 1$.
Assume that $f(b) > \cosh(b)$ for some $b > 0$. Define
$$
 a = \max \{ x \in [0, c] : f(x) \le \cosh(x) \} \, .
$$
Then $f(a) = \cosh(a)$ and $f(x) > \cosh(x) > 1$ for $a < x \le b$. Now integrate as above to obtain a contradiction.
A: Since $f(0) = 1$ and $f^2 - f'^2 ≥ 1$ we have $f'(0) = 0$
and $\cosh'(x) = 0$, Also $f(0) = cosh(0)$ so it will be enough to prove that $cosh'(x) ≥ f'(x)$ for $x > 0$. It is not so hard you can easily continue this.
