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?):

$x^2<x^2+1$, 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?

  • 1
    $\begingroup$ Consider the cases $x <0$ and $x \geq 0$. $\endgroup$ Jun 20, 2020 at 8:14
  • $\begingroup$ You can start with $x^2 < x^2+1$, and square root both sides. Make sure to remember that $\sqrt{x^2}=|x|$ and $|x| \ge x$. $\endgroup$ Jun 20, 2020 at 8:14
  • $\begingroup$ 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. $\endgroup$
    – GReyes
    Jun 20, 2020 at 8:15
  • $\begingroup$ Consider the 3 cases $x<0$,$x=0$,$x>0$ separately. $\endgroup$
    – Peter
    Jun 20, 2020 at 8:15
  • $\begingroup$ In fact the reverse implies that $|x|<\sqrt{x^2+1}$ from here it is easy to continue just consider two cases. $\endgroup$ Jun 20, 2020 at 8:16

2 Answers 2


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)$x<0$.

(in the last the inequality is obvious.)

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: $$ \underbrace {(\sqrt {x^2+1}+x)}_{\ge0}(\sqrt {x^2+1}-x)=1>0\implies x <\sqrt {x^2+1}. $$


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