Alrighty, here is my attempt at a solution. Bear with me, I'm terrible with LaTeX, so if the formatting is awful, someone please feel free to fix it.
- Building off of what @Did said.
First, let $f(x) = x^2-2x+1$
- Show that the absolute value around $f(x)$ is unnecessary (needs Calculus, but for simple proof, plot it. :)).
$\min(f(x)) \Rightarrow f'(x) = 2x-2 = 0, \Rightarrow x = 1$ ($x = 1$ is the point where the slope goes from negative to positive, or vice versa)
$f''(x) = 2 > 0 \Rightarrow$ concave upwards, and thus the only minima.
$f(1) = 0$, so $\min(f(x)) = 0$, so the $\text{abs}()$ is not necessary --- we thus remove it.
- Simplify remaining equation
We are now left with:
$\frac{|x-3|}{x} + x^2-2x+1 + x > 0 \Leftrightarrow |x-3| + x\cdot(x^2-x+1) > 0 \Leftrightarrow $ $|x-3| + x^3 -x^2 + x > 0$
This means, that there are only two possible domains that need searching, so why not search them exhaustively?
For all $x \geq 3$, we have: $|x-3| +x^3-x^2+x > 0 \Leftrightarrow x^3-x^2+2x-3 > 0.$
Therefore, to find the locations of interest, look at where that inequality equals zero.
$x^3-x^2+2x-3 = 0$
To solve this, you need to make use of the cubic formula, which is huge. Suffice it to say (aka Wolfram Alpha) gives the exact real solution as: $\frac{1}{3} \left(1-5^{2/3} \left( \frac{2}{13+3 \sqrt{21}} \right) ^ {1/3} + \left( \frac{5}{2}(13+3 \sqrt{21}) \right) ^ {2/3} \right),$ which is not what we are looking for.
Therefore, let's try the other possibility, $x < 3$: $-(x-3)+x^3-x^2+x > 0$ $x^3-x^2+3 > 0$
Same methodology: $x^3-x^2+3 = 0$
Wolfram Alpha gives: $\frac{1}{3} \left(1-\left(\frac{2}{79-9 \sqrt{77}}\right)^{1/3}-\left(\frac{1}{2} (79-9 \sqrt{77})\right)^{1/3}\right)$
Which is exactly the expression you were looking for!
Now you just have to try values on either side, and figure out which side works. Best bet is a rounded numerical approximation of the last value --- run through a calculator.