Please check my proof for $x^2+|x-2|>1$ Let $f(x)=x^2+|x-2|-1$.
If $x <2,$ then
$f(x)=x^2-x+1 \implies f'(x)=2x-1 \implies f_{min}=f(1/2)=\frac{3}{4} >1.$
If $x>2$, then $f(x)=x^2+x-3 \implies f'(x)=2x+1>0$. So the function is increasing for $x>2$
The function $f(x)$ has just one min, so $f(x)>f(1/3)=3/4  >0$, hence for all real values $x^2+|x-2| >1.$.
Is my proof correct. What can be other methods to prove this?

Ehen $x>2$ the inequation is $$x*2+x-3 >0\implies x \in (-\infty, \frac{-1-\sqrt{13}}{2}) \cup (\frac{\sqrt{13}-1}{2}, \infty)$$
So the solution is $x \in (2, \infty).$
When $x \le 2$, the inequation becomes $x^2-x+1>0$ which is true as $B^2< 4AC$. Then the solutions are $(-\infty, 2]$.
Finally, the solutions of this inequation are all real numbers, hence proved.
