Squaring a simple inequality

Since $x^2+1>0$, the inequality

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

is always satisfied for $2x-1\le 0 \implies x\le \frac12$.

Then for $2x-1>0$ we can square to obtain

$$x^2+1>4x^2-4x+1 \iff3x^2-4x<0 \iff 0<x<\frac43$$

and we need to take the solutions compatible with the condition $2x-1>0$ that is $x>\frac12$.

Therefore solutions are

• $x\le \frac12$
• $\frac12<x<\frac43$

that is $x<\frac43$.

It's obvious that for $x\leq0$ our inequality is true.

Let $x>0$.

Thus, we need to solve $$\sqrt{x^2+1}-x>x-1$$ or $$\frac{1}{\sqrt{x^2+1}+x}>x-1.$$ Now, consider two functions: $f(x)=\frac{1}{\sqrt{x^2+1}+x}$ and $g(x)=x-1$.

We see that $f$ decreases for $x>0$ and $g$ increases, which says that graphs of $f$ and $g$ intersect in one point maximum.

But $f\left(\frac{4}{3}\right)=g\left(\frac{4}{3}\right)$, which gives the answer: $$\left(-\infty,\frac{4}{3}\right).$$