How do I prove the maximum of this function I have the function
$$y = x - \sqrt{x^2 - 1}$$
which must have a maximum of $1$ at $x = 1$, as after that you're taking $x$ and subtracting something slightly smaller than $x$, tending to $0$ as $x$ tends to infinity, however its derivative of 
$$1 - \frac{x}{\sqrt{x^2 - 1}}$$
is undefined at $x = \pm 1$, as is its second derivative. 
How can I prove this function is bounded above by 1, and that the absolute value of y doesn't exceed 1 at some point 0 < x < 1?

To get maximum of this function, clearly, $x \ge 0$. So we have:
$$x-\sqrt{x^2-1} = x-\sqrt{x^2-1} \times \frac{x+\sqrt{x^2-1} }{x+\sqrt{x^2-1} } = \frac{1 }{x+\sqrt{x^2-1} } \ge \frac{1}{1+\sqrt{1^2-0}}= 1,$$
Note that $x+\sqrt{x^2-1}$ is an increasing function at $[1, +\infty]$.
