<s>[INST] is $y = \sqrt{x^2 + 1}− x$ a injective (one-to-one) function? I have a function  $y = \sqrt{x^2 + 1}− x$ and I need to prove if it's a Injective function (one-to-one).  The function f is injective if and only if for all a and b in A, if f(a) = f(b), then a = b
$\sqrt{a^2 + 1} − a = \sqrt{b^2 + 1} − b$
$\sqrt{a^2 + 1} + b = \sqrt{b^2 + 1} + a$
$(\sqrt{a^2 + 1} + b)^2 = (\sqrt{b^2 + 1} + a)^2$
$a^2 + 1 + b^2 + 2b\sqrt{a^2 + 1} = b^2 + 1 + a^2 + 2a\sqrt{b^2 + 1}$
$2b\sqrt{a^2 + 1} = 2a\sqrt{b^2 + 1}$
$b\sqrt{a^2 + 1} = a\sqrt{b^2 + 1}$
$(b\sqrt{a^2 + 1})^2 = (a\sqrt{b^2 + 1})^2$
$b^2(a^2 + 1) = a^2(b^2 + 1)$
$b^2a^2 + b^2 = a^2b^2 + a^2$
$b^2 = a^2$
$\sqrt{b^2} = \sqrt{a^2}$
$b = \pm a$
then the function is not injective because $b$ in not equal to $a$, but using online solver the function is injective.
 [/INST] Hint:
$$y=\sqrt{x^2+1}-x=\frac1{\sqrt{x^2+1}+x}$$
Then $$f(a)=f(b) \Leftrightarrow \frac1{\sqrt{a^2+1}+a}=\frac1{\sqrt{b^2+1}+b}\Leftrightarrow$$
$$\Leftrightarrow \sqrt{a^2+1}+a=\sqrt{b^2+1}+b \Leftrightarrow a=b$$
$f(t)=\sqrt{t^2+1}+t -$ strictly increasing
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