MHB How Do You Calculate the Total Length of Intervals for x in This Inequality?

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The inequality $\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge 1$ defines a set of real numbers $x$ that results in intervals of the form $a<x\le b$. Participants are tasked with calculating the total length of these intervals. The discussion includes a reference to the Problem of the Week (POTW) guidelines for proper submission. A suggested solution is available for review, but no responses have been provided for this particular problem yet. The focus remains on determining the sum of the lengths of the identified intervals.
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Here is this week's POTW:

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The set of real numbers $x$ for which

$\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge 1$

is the union of intervals of the form $a<x\le b$.

Find the sum of the lengths of these two intervals.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered POTW #402.(Sadface)

You can find the suggested solution by other below.
Let $f(x)=\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}$ .

Note that

$f(x)-f(y)=(y-x)\left(\dfrac{1}{(x-2009)(y-2009)}+\dfrac{1}{(x-2010)(y-2010)}+\dfrac{1}{(x-2011)(y-2011)}\right)$

If $x<y<2009$, then

$y-x>0$,

$\dfrac{1}{(x-2009)(y-2009)}>0,\,\dfrac{1}{(x-2010)(y-2010)}>0,\,\dfrac{1}{(x-2011)(y-2011)}>0$

Thus $f$ is decreasing on the interval $x<2009$, and because $f(x)<0$ for $x<0$, it follows that no values $x<2009$ satisfy $f(x)\ge 1$.

If $2009<x<y<2010$, then $f(x)-f(y)>0$ as before. Thus $f$ is decreasing in the interval $2009<x<2010$. Moreover $f\left(2009+\dfrac{1}{10}\right)=10-\dfrac{10}{9}-\dfrac{10}{19}>1$ and $f\left(2010-\dfrac{1}{10}\right)=\dfrac{10}{9}-10-\dfrac{10}{11}<1$. Thus there is a number $2009<x_1<2010$ such that $f(x)\ge 1$ for $2009<x\le x_1$ and $f(x)<1$ for $x_1<x<2010$.

Similarly, $f$ is decreasing on the interval $2010<x<2011$, $f\left(2010+\dfrac{1}{10}\right)>1$, and $f\left(2011-\dfrac{1}{10}\right)<1$. THus there is a number $2010<x_2<2011$ such that $f(x)\ge 1$ for $2010<x\le x_2$ and $f(x)<1$ for $x_2<x<2011$.

Finally, $f$ is decreasing on the interval $x>2011$, $f\left(2011+\dfrac{1}{10}\right)>1$ and $f(2014)=\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{3}<1$. Thus there is a number $x_3>2011$ such that $f(x)\ge 1$ for $2011<x\le x_3$ and $f(x)<1$ for $x>x_3$.

The required sum of the lengths of these three intervals is

$x_1-2009+x_2-2010+x_3-2011=x_1+x_2+x_3-6020$

Multiplying both sides of the equation $\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}=1$ by $(x-2009)(x-2010)(x-2011)$ and collecting terms on one side of the equation gives

$x^3-x^2(2009+2010+2011+1+1+1)+ax+b=0$

where $a$ and $b$ are real numbers. The three roots of this equation are $x_1,\,x_2$ and $x_3$. Thus $x_1+x_2+x_3=6020+3$ and consequently the required sum equals 3.
 
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