Difficult inequality, difficult solution, but how to prove?

Having $$\frac{|x-3|}{x} + |x^2-2x+1| + x > 0$$ how can I arrive to the solution:

$$x < \frac 13 \left( 1-\sqrt[3]{\frac{2}{79-9\sqrt{77}}} - \sqrt[3]{\frac 12 \left(79 - 9\sqrt{77}\right)} \right)$$

?

Of course, $x \neq 0$, and then I've tried all the possible ways I could think of by splitting up the real numbers in three intervals, $x < 0$, $(x>0 \land x<3)$ and $x > 3$, but I couldn't come up with anything as close to the solution above.

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I would start from $|x^2-2x+1|=x^2-2x+1$. (But I do not understand your solution since the inequality is obviously satisfied for every $x\gt0$.) –  Did Nov 5 '12 at 12:03
Yeah, I've done that, but wolfram alpha says that's the solution... –  Flavius Nov 5 '12 at 12:11
The consequences (regarding the proper attitude to adopt with respect to W|A results) seem obvious. –  Did Nov 5 '12 at 12:15
Persuing did's remark, take cases on whether $x>3$ or $x<3$ and "multiply through" by $x$ to get cubic expressions on the left side. Also need to consider the sign of $x$ here, at the "multiply by $x$" step. At any rate there will definately be cubic equation(s) to solve, so some expression like your answer might appear. –  coffeemath Nov 5 '12 at 12:15
Yeah, but if I split it in cases like that, I'll end up with different intervals (to unify). How to go about to end up with a single inequation? –  Flavius Nov 5 '12 at 12:20
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