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

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.

• $\sqrt {x^{2}+1} >\sqrt {x^{2}}=|x|\geq -x$. Mar 19, 2022 at 11:36
• @KaviRamaMurthy Thank you so much! This is exactly what I was missing Mar 20, 2022 at 0:34

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

• For benefit of OP - note how this follows the (familiar?) pattern $(a+b)(a-b)=a^2-b^2$ where $a$ and $b$ may be expressions involving square roots. When you first see such things they can be confusing or seem mysterious, but once understood they can come in handy. Mar 19, 2022 at 12:26

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 $$\sqrt{x^2 + 1} > - x$$ $$\sqrt{x^2 + 1} + x >0.$$

WLOG

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

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

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.

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