Prove: $\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x}=0$ has a real root between $\frac13a$ and $\frac23a$, and one between $-\frac23b$ and $-\frac13b$ 
If $a$ and $b$ are positive numbers, prove that the equation
  $$\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x}=0$$
  has two real roots; one between $\frac13a$ and $\frac23a$, and one between $-\frac23b$ and $-\frac13b$.


My attempt:
I simplified the given expression, applied the discriminant for two real roots and got stuck on this inequality: $4({a}^2 - {ab}+{b}^2)$$>$$0$
Any help would be most appreciated.
 A: $$4(a^{2}- ab + b^{2}) = (2a-b)^{2} + 3b^{2}$$

I think question is wrong. For exmaple, if $a = 1, b = 2$, then two solutions are $1 \pm \frac{1}{\sqrt{3}}$, and none of them lies between $-4/3$ and $-2/3$ since both are positive.
A: Hui is right the equation should be $$ f(x)=\frac{1}{x-a}+\frac{1}{x+b}+\frac{1}{x} =0 ~~~a,b >0. ~~~(1)$$
Then $$f(a/3)=\frac{9(a+b)}{2a(a+3b)}>0$$ and $$f(2a/3)=~-\frac{9b}{2a(2a+3b)}<0.$$ So by the Intermediate Value Theorem (IVT) this equation (1) has one root in $(a/2,2a/3).$
Next, $$ f(-b/3)=~-\frac{9(a+b)}{2b(3a+b)}<0$$ and $$f(-2b/3)=\frac{9a}{2b(3a+2b)}>0.$$ So again by IVT, equation (1) has one root in $(-2b/3,-b/3).$
Hence proved.
A: Note: The defining equation is symmetric in $a$ and $b$; the ostensible root-bounding intervals are not. This, in and of itself, doesn't guarantee that the question is invalid, but it reasonably raises some suspicions. Be that as it may ...

Clearing fractions gives the quadratic
$$3 x^2 - 2 (a+b) x + a b = 0$$
which we can solve explicitly:
$$x = \frac13\left(\;a+b \pm \sqrt{a^2-a b+ b^2}\;\right)$$
Presumably, from the target root-bounding intervals, one of the roots "should be" positive, and one negative. The positive root must arise from taking "$\pm$" to be "$+$", so that we have 
$$\frac13a \leq \frac13\left(\;a+b + \sqrt{a^2-a b+ b^2}\;\right) \leq \frac23a \quad\to\quad 0 \leq b + \sqrt{a^2-a b+ b^2} \leq a \quad\to\quad b\leq a$$
This condition on $a$ and $b$ was not assumed, so the conclusion is not valid. (Likewise, having the negative root bounded by the $b$-terms introduces an additional condition.) $\square$
