# Prove $x<\sqrt{x^2+1}$.

Prove $$x<\sqrt{x^2+1}$$.

I am pretty sure this an easy question as the inequality seems obviously true, but I am not entirely convinced by my argument.

So I squared both sides (is this allowed?):

$$x^2, so $$0<1$$ so the inequality is obviously true.

However, I am unconvinced that this process is reversible due to the squaring, so could someone just explain whether this is correct?

• Consider the cases $x <0$ and $x \geq 0$. Jun 20, 2020 at 8:14
• You can start with $x^2 < x^2+1$, and square root both sides. Make sure to remember that $\sqrt{x^2}=|x|$ and $|x| \ge x$. Jun 20, 2020 at 8:14
• If $x<0$ your inequality is obvious, since the right hand side is $\ge 0$. If $x\ge 0$ your squaring is correct, the squared inequality is equivalent to the original one and your proof works. Jun 20, 2020 at 8:15
• Consider the 3 cases $x<0$,$x=0$,$x>0$ separately. Jun 20, 2020 at 8:15
• In fact the reverse implies that $|x|<\sqrt{x^2+1}$ from here it is easy to continue just consider two cases. Jun 20, 2020 at 8:16

You can not use squaring of the both sides directly because $$x$$ can be negative.
If so, you need to consider two cases: 1)$$x\geq0$$ and 2)$$x<0$$.
I think it's better to use a way without squaring:$$\sqrt{x^2+1}>\sqrt{x^2}=|x|\geq x.$$
By definition $$\sqrt {x^2+1}\ge0$$. If $$x <0$$, then $$x <0\le \sqrt {x^2+1} \implies x <\sqrt {x^2+1}$$. If $$x\ge0$$ we have: $$\underbrace {(\sqrt {x^2+1}+x)}_{\ge0}(\sqrt {x^2+1}-x)=1>0\implies x <\sqrt {x^2+1}.$$