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: 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.
A: As an alternative we have that
$$x^2+|x-2|>1\iff x^2-1>-|x-2|$$
which is trivially true for $x<-1$ and $x>1$ then for $|x|\le1$ we obtain
$$x^2-x+1>0$$
which is true since $\Delta<0$.
A: The derivative of $|x-2|$ is $\pm1$, depending on the sign of $x-2$. Hence the extrema can be located at
$$2x=\pm1.$$ The root $x=-\dfrac12<2$ must be rejected, and the global minimum is
$$\frac1{2^2}+\left|\frac12-2\right|>1.$$

To address David's objections, notice that the function is differentiable for all $x\ne 2$, and continuous at $f(2)>1$, so the computation of the global minimum by differentiation makes sense.
A: 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.
