<s>[INST] Different solutions with different results for an inequality Find m such that the following inequality:
$$\left|4x-2m-\frac{1}{2}\right| > -x^2 +2x + \frac{1}{2} - m$$
is always true for $\forall x \in R$.
1st solution:
1st case
$$4x-2m-\frac{1}{2} > -x^2 + 2x +\frac{1}{2} -m$$
$$<=>x^2+2x-m-1>0$$
$$\Leftrightarrow 1^2+(m+1)< 0$$
$$\Leftrightarrow m<- 2$$
2nd case
$$4x-2m-\frac{1}{2}< -(-x^2 + 2x +\frac{1}{2} -m)$$
$$\Leftrightarrow x^2-6x+3m>0$$
$$\Leftrightarrow 3^2-3m<0$$
$$\Leftrightarrow m>3$$
2nd solution:
The inequality is the same as:
$$(x-1)^2+|4x-2m-\frac{1}{2}|>\frac{3}{2}-m$$
Since the left-hand side is always positive, in order for the inequality to be always true, $\frac{3}{2}-m$ has to be negative, or $m > \frac{3}{2}$

The 2 solutions give different answers, so I was quite confused
But I get more confused as Wolfram Alpha gives me the solution:
$$m > \sqrt{3} - \frac{1}{4} \text{  or  } m < -\sqrt{3} - \frac{1}{4} $$
There's a high chance that Wolfram Alpha's solution is correct (after testing out some $m$ value). How do I approach their solution? (Or maybe if you believe that solution is wrong, then what's the exact solution to the problem?)
 [/INST] Your two solutions are mathematically correct, but you when you reach $m<-2$, you have calculated for which values of $m$, we have $m+2>0$ and you have not shown that $\forall m>-2, (x+1)^2>m+2, \forall x \in R$. This explain the error in the first method. For the second method, you have shown that the inequality:
$$(x-1)^2+|4x-2m-\frac{1}{2}|>\frac{3}{2}-m$$
have solutions when $m<\frac{3}{2}$, but this not explain for which $m$, we have:
$$\left|4x-2m-\frac{1}{2}\right| > -x^2 +2x + \frac{1}{2} - m$$
In general, when you have to solve this type of inequality, you have to solve the system:
$$\left\{\begin{matrix}
4x-2m-\frac{1}{2}\geq0
\\ -x^2 +2x + \frac{1}{2} - m\geq0
\\ 4x-2m-\frac{1}{2} > -x^2 +2x + \frac{1}{2} - m
\end{matrix}\right.
\vee
\left\{\begin{matrix}
4x-2m-\frac{1}{2}<0
\\ -x^2 +2x + \frac{1}{2} - m<0
\\ -(4x-2m-\frac{1}{2}) > -x^2 +2x + \frac{1}{2} - m
\end{matrix}\right.
$$
These system are quite complicated, and I will give directly the solutions:
$$\left\{\begin{matrix}
x>\sqrt{m+2}-1
\\ x<-\sqrt3\sqrt{3-m}+3
\end{matrix}\right.$$
When $m<-2$ and $m>3$ the square roots are not defined, so these values of $m$ are correct. 
Also, we want the two expression to be the same because we want a single solution, so:
$$\sqrt{m+2}-1=-\sqrt3\sqrt{3-m}+3$$
The solutions are:
$$m=\sqrt3-\frac{1}{4} \vee m=-\sqrt3-\frac{1}{4}$$
When $-\sqrt3-\frac{1}{4}<m<\sqrt3-\frac{1}{4}$ there are solutions, while when:
$$m > \sqrt{3} - \frac{1}{4} \lor m < -\sqrt{3} - \frac{1}{4}$$
there aren't.
</s>