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?

A standard way to prove such an inequality without derivatives would be to complete the square.
For $x \ge 2$ the case is simpler:
$$x^2 + |x-2| \ge 4 + 0 > 1$$
For $x < 2$ we have $|x - 2| = 2-x$. Hence:
$$\begin{align}x^2 + |x-2| &= x^2-x+2 \\&= x^2-x+\frac14 + \frac74 \\&= \left(x-\frac12\right)^2+\frac 74 \\&\ge \frac74>1\end{align}$$
using the fact that squares are nonnegative.
