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

Question

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?

Answer 1

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:$$\sqrt{x^2+1}>\sqrt{x^2}=|x|\geq x.$$

Answer 2

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