|x²-2x| + |x-4| > |x²-3x+4| , How do I solve for all real x? How do I solve this for all real x?
|x²-2x| + |x-4| > |x²-3x+4|
Looking at the question it is clear that it states |a| + |b| > |a-b|. How to proceed?
 A: We know that by the triangle inequality
$$|a|+|b|\geq|a+b|$$ and the equality occurs for $ab\geq0.$
Now, we have $$|x^2-2x|+|x-4|=|x^2-2x|+|4-x|\geq|x^2-3x+4|.$$
Id est, our inequality is equivalent to   $$(x^2-2x)(4-x)<0$$ or
$$x(x-2)(x-4)>0,$$ which by the intervals method gives $$(4,+\infty)\cup(0,2).$$
A: First, note that $x^2-3x+4$ is always positive because discriminant $D=(-3)^2-4\times4<0$.
Case 1: $x\in(-\infty, 0]$
In this range: $|x^2-2x|=x^2-2x$, $|x-4|=-(x-4)$ and the inequality becomes:
$$x^2-2x-(x-4)>x^2-3x+4$$
$$x^2-3x+4>x^2-3x+4$$
...which is never true. So the ineqality has no solution in the given range.
Case 2: $x\in(0, 2]$
In this range: $|x^2-2x|=-(x^2-2x)$, $|x-4|=-(x-4)$ and the inequality becomes:
$$-(x^2-2x)-(x-4)>x^2-3x+4$$
$$-x^2+x+4 > x^2-3x+4$$
$$0 > 2x^2-4x$$
$$0 > x^2-2x=x(x-2)$$
...and this is obviously true for $x\in(0,2)$. Combined with the initial condition that $x\in(0,2]$ this gives one part of the solution: $x\in(0,2)$
Case 3: $x\in(2, 4]$
In this range: $|x^2-2x|=x^2-2x$, $|x-4|=-(x-4)$ and the inequality becomes:
$$x^2-2x-(x-4)>x^2-3x+4$$
$$x^2-3x+4>x^2-3x+4$$
...which is never true. 
Case 4: $x\in(4, +\infty)$
In this range: $|x^2-2x|=x^2-2x$, $|x-4|=x-4$ and the inequality becomes:
$$x^2-2x+x-4>x^2-3x+4$$
$$-x-4>-3x+4$$
$$2x-8>0$$
$$x>4$$
So all values of $x$ in the given range are also solutions to the inequality.
If you summarize all four cases you get the final result:
$$\boxed{x\in(0,2)\cup(4,+\infty)}\tag{1}$$
The solution can be confirmed by looking at the plot of the function: 
$$f(x)=|x^2-2x| + |x-4| - |x^2-3x+4|$$

...which is positive for values of $x$ defined by (1).
