Solving $|x^2-2x|+|x-4|>|x^2-3x+4|$ 
How do I solve $|x^2-2x|+|x-4|>|x^2-3x+4|?$

I can see the difference of the two terms on the left gives the term on the right. Now,what should I do? Is there any general method for solving $$|a|+|b|>|a-b|?$$
Thanks for any help!!
 A: Note that $|x^2-2x|+|x-4|=|x^2-2x|+|-x+4|\ge|x^2-3x+4|$ with the equality holds if and only if $x^2-2x$ and $-x+4$ are of the same sign. 
Therefore, $|x^2-2x|+|x-4|>|x^2-3x+4|$ if and only if $x^2-2x$ and $x-4$ are of the same sign. So we have
\begin{align*}
(x^2-2x)(x-4)&>0\\
x(x-2)(x-4)&>0\\
0<x<2 \quad\textrm{or}\quad x&>4
\end{align*}
A: rewring as
$$|x||x-2|+|x-4|>x^2-3x+4$$ since $$x^2-3x+4>0$$ for all real $x$
we distinguish four cases:
$$x\geq 4$$
$$2\le x<4$$
$$0\le x<2$$
$$x<0$$
$$x(x-2)+x-4>x^2-3x+4$$
$$x(x-2)-x+4>x^2-3x+4$$
$$x(-x+2)-x+4>x^2-3x+4$$
$$-x(-x+2)-x+4>x^2-3x+4$$
Can you finish?
Finally we find: $$0<x<2$$ or $$x>4$$
A: Solving this is just a matter of using the definition of |x|.
This is |x||x- 2|+ |x+ 4|> |x^2- 3x+ 4|.
x^2- 3x+ 4 is positive for all x so we need to consider three cases:
x< -4.  Then all of x, x- 2, and x+ 4 are negative.  This is (-x)(2- x)- (x+ 4)= x^2- 3x- 4> x^2- 3x+ 4. That reduces to -4> 4 which is never true. There is no x< -4 that satisfies this.
-4< x< 0.  x+ 4 is positive but x and x+ 2 are still negative so this is (-x)(2- x)+ (x+ 4)= x^2- x+ 4> x^2- 3x+ 4.  That reduces to 2x> 0 but this case is for x< 0 so there is no x between -4 and 0 that satisfies this.
0< x< 2.  Now both x+ 4 and x are positive while x- 2 is negative so this is (x)(2- x)+ (x+ 4)= -x^2+ 3x+ 4> x^2- 3x+ 4.  That reduces to -2x^2+ 6x= x(6- 2x)> 0.  A product is positive if both numbers are positive or if both numbers are negative.  Either x> 0 and 6> 2x so x> 3 or x< 0 and 6< 2x.  The first is satisfied for x> 3 and the second for x< 0.  Neither of those is true for 0< x< 2 so there is no x that satisfies this.
2< x.  Now all three terms are positive so this is (x)(x- 2)+ x+ 4= x^2- x+ 4> x^2- 3x+ 4.  That reduces to 2x> 0 or x> 0. Since, here, x> 2 x is also greater than 0. This inequality is satisfied for x> 2..
