Prove that given $ a < b < c $ this equation: $$\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0 $$
has precisely 2 real roots.
I understand there are 3 point of discontinuities, but I have no idea how to prove this. Can you give me a hint?
Thanks in advance.
Let $f(x)=(x-a)(x-b)(x-c)$. Then
$$\frac{f'(x)}{f(x)}=\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c} \,.$$
Thus, the problem asks you to prove that $f'(x)=0$ has exactly two real roots not equal to $a,b,c$.
Since $f(x)$ is cubic, $f'$ is quadratic, thus it has at most two real rules.
By Rolle Theorem, $f'$ has a root in $(a,b)$ and a root in $(b,c)$.
HINT:
Rearrange the equation to get $3x^2-2(a+b+c)x+ab+bc+ca=0$
$$\text{The discriminant }\{2(a+b+c)\}^2-4\cdot3\cdot(ab+bc+ca)$$
$$=4(a^2+b^2+c^2-ab-bc-ca)$$
$$=4\cdot\frac{\{(a-b)^2+(b-c)^2+(c-a)^2\}}2>0 \text{ for distinct real }a,b,c$$
What can we conclude from this?
Here's what some of the graphs look like. This is not a proof but, perhaps, it will give you some clue as to what's going on, particularly if you can show that the generic graph looks like this. Thus, you might think of this as the geometric motivation behind some of the other answers that encourage you to think about where the function behaves in some fashion.
Hint: Make a common denominator for the three factors. Then solve the resulting equation (or check the value of its discriminant).
To be a little bit more explicit, bring it into the following form: $\frac{(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)}{(x-a)(x-b)(x-c)}=0$. Now simplify and find the discriminant (or solve explicitly).
Hints:
Look at the derivative of the function to determine where it's increasing/decreasing.
Consider the behaviour near the poles.
This should be enough to solve the problem.
Set $y=x-c$, so that $x=y+c$ and $x-a=y+c-a$, $x-b=y+c-b$. Set $r=c-a$ and $s=c-b$. Then your equation becomes $$ \frac{1}{y+r}+\frac{1}{y+s}+\frac{1}{y}=0 $$ or $$ y(y+s)+y(y+r)+(y+r)(y+s)=0. $$ Simplifying you get $$ 3y^2+2(r+s)y+rs=0 $$ and the reduced discriminant is $$ (r+s)^2-3rs=r^2-rs+s^2>0 $$ by well known facts about second degree polynomials.
Now $r>0$ and $s>0$, so $0$ is not a root. If we substitute $-r$ to $y$ we get $$ 3r^2-2r^2-rs+rs=r^2\ne0 $$ and similarly $-s$ is not a root. Thus both roots of the polynomial in $y$ are solutions of the equation in $y$ and therefore the original equation has two real solutions.
Consider the 4 regions: for $x>c$ all terms are positive, hence there is no real root there. For $x \to c^{-}$, it tends to $-\infty$, for $x\to b^{+}$ it tends to $+\infty$, so there must be a real root in $(b,c)$. Make this argument rigorous.
It can also be done by contradiction.
HINT: If possible,suppose there is at most one real root.Then other $2$ roots must be conjugate in pairs. Let other roots be of the form $\alpha +i \beta$ and $\alpha -i \beta$. Now put $x=\alpha +i \beta$ and $x=\alpha -i \beta$ in the given equation and see what happens ....
$$\text{If }f(x)=\sum_{1\le r\le n}\frac1{x-a_r}$$ where $a_i$s are real,
If one of the root of $f(x)=0$ is $p+iq,$
$$0=f(p+iq)=\sum_{1\le r\le n}\frac1{p+iq-a_r}\ \ \ \ (1)$$
Using Complex conjugate root theorem, $p-iq$ must be another root
$$\implies 0=f(p-iq)=\sum_{1\le r\le n}\frac1{p-iq-a_r}\ \ \ \ (2)$$
$$\text{Now, }\frac1{p-iq-a_r}-\frac1{p+iq-a_r}=\frac{2iq}{(p-a_r)^2+q^2}$$
$$(2)-(1)\implies 2iq\sum_{1\le r\le n}\frac1{(p-a_r)^2+q^2}=0\implies q=0$$
So, the roots of $f(x)$ can not have imaginary parts.
Now, we know, every polynomial equation having complex coefficients and degree $m\ge1$ has exactly $m$ roots.