|x²-4|+|x|=|x²+x-4|
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Note that |x2−2x|+|x−4|=|x2−2x|+|−x+4|≥|x2−3x+4| with the equality holds if and only if x2−2x and −x+4 are of the same sign.
Therefore, |x2−2x|+|x−4|>|x2−3x+4| if and only if x2−2x and x−4 are of the same sign. So we have
(x2−2x)(x−4)x(x−2)(x−4)0<x<2orx>0>0>4
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