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?)

Your mistake in the first solution: at some stage you drop the variable $x$ by using $\forall x\in\mathbb R$. But as you are doing this case analysis, the $\forall$ no longer holds.
Your mistake in the second solution: from $a>0, a>b$, you conclude $0>b$, which is not tight. (Take $a=2,b=1$.)
