Given $\frac{1}{x}+\frac{1}{x-a}+\frac{1}{x-b}=0$, $\{a,b\}\subset \mathbb R^+$, show the roots are real and within defined intervals 
Given: $\{a,b\}\subset \mathbb R^+$, $P(x)=\frac{1}{x}+\frac{1}{x-a}+\frac{1}{x-b}=0.$ (correction: last term is $\frac{1}{x{\color{red}+}b})$
Show: all roots of $P(x)$ are real, one in the interval $(-b,0)$ and the other in the interval $(0,a)$.

Question asked in an entrance exam.
My attempt: developing $P(x)$, we get to
$$x^2-\frac{2}{3}(a+b)x+\frac{ab}{3}=0$$
with discriminant defined by
$$\Delta=\frac{4}{9}(a+b)^2-\frac{4}{3}ab=\frac{4}{9}((a+b)^2-3ab)=\frac{4}{9}(a^2+b^2-ab)$$
As it is easy to show that $a^2+b^2-ab\ge 0$ it follows that $\Delta\ge 0$. This solves the first part of the question.
I tried, unsuccessfully, to solve the second part, which asked to show that the roots are within two specific intervals. Hints and full answers are welcome.
Edit 1. The term $\frac{1}{x-b}$ is, most likely, $\frac{1}{x+b}$. See comments below for an explanation.
 A: I am using calculus. Please tell me if it is not allowed in the exam, then I can post an alternate solution.
Case $1$ : $P(x) = \frac 1{x} + \frac 1{x-a} + \frac 1{x-b} = 0$ (This was done before the edit, but is included for completeness)
$P(x)$ is a continuous function on $\mathbb R \backslash\{0,a,b\}$. Without loss of generality, let us assume that $0 < a < b$. Then, in the interval $(0,a)$,  note that as $x \to 0^+$, $P(x)$ eventually becomes positive, and as $x \to a^-$, $P(x)$ eventually becomes negative, and therefore, by the intermediate value theorem, $P(x)$ has a root in $(0,a)$.
Similarly, in the interval $(a,b)$,  note that as $x \to a^+$, $P(x)$ eventually becomes positive, and as $x \to b^-$, $P(x)$ eventually becomes negative, and therefore, by the intermediate value theorem, $P(x)$ has a root in $(a,b)$.
Finally, we note that for $x < 0$, all three quantities $\frac{1}{x},\frac{1}{x-a},\frac{1}{x-b}$ are strictly negative, hence their sum is strictly negative. Similarly, for $x>b$, the sum is strictly positive. Hence, no roots lie in these regions.
Furthermore, $P(x)$ has no complex roots. To see this, note that by taking the common denominator, $$\frac 1{x} + \frac 1{x-a} + \frac 1{x-b} = \frac{3x^2+x(-2a-2b)+ab}{x(x-a)(x-b)}$$
Therefore, $P(x)=0$ precisely when $3x^2 + x(-2a-2b) + ab = 0$. This can have at most two roots in the entire complex plane because it has degree $2$. However, we have already located those roots above, showing that there are no other roots available.
Therefore , there are precisely two roots of this equation, both real.
Case $2$ : $P(x) = \frac 1x + \frac 1{x-a} + \frac 1{x+b} = 0$.
We will use calculus, yet again. Now, we have the following regions to consider :

*

*$x<-b$ : Each of $\frac 1{x},\frac{1}{x-a},\frac 1{x+b}<0$, therefore there is no root in this interval.


*$-b<x<0$ : As $x \to -b^+$, $P(x) \to +\infty$, while as $x \to 0^-$, $P(x) \to -\infty$. Therefore, by the intermediate value theorem, there is a root in $(-b,0)$.


*$0<x<a$ : As $x \to 0^+$, $P(x) \to +\infty$, while as $x \to a^-$, $P(x) \to -\infty$. Therefore, by the intermediate value theorem, there is a root in $(0,a)$.


*$x>a$ : Each of $\frac 1x , \frac 1{x-a},\frac{1}{x+b}$ is positive in this interval, hence there is no root.
However, similar to the argument in Case $1$, $P(x)$ has only two roots in total : and we found both above. Hence, these are the roots of the equation. Furthermore, if one needs to find them, then one can use the quadratic formula on the equation derived as the numerator of $P(x)$ after one takes the common denominator.
A: I think the problem is not true. When a is smaller than b, consider the quadratic function you get , let x equal to 0, a and b, you will know that one root lies between 0 and a, another root lies between a and b. When a is bigger than b, it is similar.
