# 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

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.

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I suppose, your first root (with $\sqrt{249}$) isn't correct. It is the root for $x^3-x^2+x-2$, not for $x^3-x^2-2x-3$. –  sas May 27 '14 at 1:39
Fixed the issue. Thanks for catching it. Looking at your LaTeX formatting code allowed me to format the right answer correctly. Thanks a ton for fixing the formatting! –  FundThmCalculus May 27 '14 at 1:49
$x^2-2x+1=(x-1)^2$ is a complete square, so you can drop the complicated curve discussion. Your first step is only true for $x> 0$, but as said several times in the comments, in this case the inequality is trivially true. The only interesting case is $x<0$, but then you have to switch the relation from $...>0$ to $...<0$ when multiplying with $x$. –  LutzL May 27 '14 at 20:35
Means, the "first step of the simplification". There is exactly one negative root of $x^3-x^2+3$ by counting sign variations in the coefficient sequence $(-1,-1,3)$ and $x^3-x^2+3<0$ is obviously true for all $x$ that are smaller than this negative root. To get an idea of the size of the expression, note that $9\sqrt{77}=\sqrt{77\cdot 81}=\sqrt{79^2-4}$. –  LutzL May 27 '14 at 20:56