Proving that $\sqrt{x^2+1}+x>0$ for all $x$

Question

Recently while dealing with inverse hyperbolic functions, I came across the expression $$\sinh^{-1}x=\ln(x+\sqrt{x^2+1})$$

We know that $f(x)=\sinh^{-1}x$ is defined for all real values of $x$ since the range for $\sinh x$ is all real numbers.

But I would like to prove this by proving that $\ln(x+\sqrt{x^2+1})$ is defined for all real values. For this, clearly, $x+\sqrt{x^2+1}>0$ but how do you prove this?

I can see that for $x>0,$ $$\sqrt{x^2+1}>\sqrt{x^2}=x>0$$ $$\implies f(x)=x+\sqrt{x^2+1}>0 \text{ }[\because x>0]$$

But what about when $x<0?$ I understand that $\sqrt{x^2+1}$ will always be greater than $0$ but how do you deal with the $+x?$

Another idea I have is to consider $|x-\sqrt{x^{2}+1}|$ and use the fact that $0<|x-\sqrt{x^{2}+1}|<1$ to show that $\sqrt{x^{2}+1}$ and $x$ cannot be "too far apart" and since $\sqrt{x^{2}+1}>1$, $x+\sqrt{x^{2}+1}>0$ for all real $x$. But this method doesn't seem very robust, so I was wondering if there was a more convincing argument and less roundabout way of doing this.

PS – I may be missing some simple algebra, but am only in high school, so would appreciate it if any answers are not too convoluted.

Answer 1

Note that the inequality is obvious when $x$ is zero or positive. It remains for us to check that $x$ is a negative number.

For any negative $x$, $|x|=-x$. Hence, $$\sqrt{x^2 + 1} > \sqrt{x^2} = |x| = -x.$$ We then have $$\sqrt{x^2 + 1} > - x$$ $$\sqrt{x^2 + 1} + x >0.$$

Answer 2

WLOG

$x=\cot2y, 0<2y<\pi$ using Principal values

$$\sqrt{x^2+1}+x=\csc2y+\cot2y=\cdots=\cot y>0$$ as $0<y<\dfrac\pi2$ and $\cos y\ne0$

Answer 3

$$(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x)=1$$ One of the factors is positive because either $x$ or $-x$ is positive, so the other factor is positive as well.

Answer 4

1)$x\ge 0$; the inequality is satisfied.

$2)x<0;$

Rewrite as

$\sqrt{x^2+1} - \sqrt{x^2}=$

$\dfrac {1} {\sqrt{x^2+1}+\sqrt{x^2}}>0,$ and we are done.

Answer 5

1. x = 0; the inequality is satisfied 1 > 0
2. x < 0; $\sqrt{x^2+1}-x > 0;$
3. x > 0; $\sqrt{x^2 + 1}+x > 0;$