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I need help proving that the next equation has only two roots (under $\mathbb R$) $$\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c} =0$$ $$a\lt b\lt c$$

Here is what I tried:

If I define a function $f(x)=\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}$ I could show that this is a continuous function and for different values I get positive or negative values and by the continuity it will be equal 0 exactly twice.

Maybe it has something to do with the function derivative?

Any ideas?

segevp
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    If $f(x)=(x-a)(x-b)(x-c)$ then $\frac{f'(x)}{f(x)} = $ that thing. $f$ vanishes at $a,b,c$ use Rolle to get that $f'$ does vanish at least twice. Use its degree to show that no more than twice. – Bettybel Jul 07 '17 at 20:52
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    "I could show that this is a continuous function" No. This is not a continuous function on ${\bf R}$. –  Jul 07 '17 at 20:53
  • Multiply by $(x-a)(x-b)(x-c)$ and solve the quadratic. Note that the reduced discriminant is$a^2 +b^2 + c^2 - a b - b c - a c \gt 0,$. – dxiv Jul 07 '17 at 20:54
  • @Jack yeah true, that was very stupid of me, – segevp Jul 07 '17 at 20:54
  • no this discriminat is $$\geq 0$$ – Dr. Sonnhard Graubner Jul 07 '17 at 21:00
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    @Dr.SonnhardGraubner Assuming you refer to my previous comment, the discriminant is $0$ iff $a=b=c$. In this case it is given that $a \lt b \lt c$ so the discriminant is strictly $,\gt 0,$, as written. – dxiv Jul 07 '17 at 21:02
  • yes ok you have right, but the condition $$a<b<c$$ is not naturally – Dr. Sonnhard Graubner Jul 07 '17 at 21:17
  • You can have a look also at this slightly different version, with any number of roots, but the principle is the same : https://math.stackexchange.com/questions/2275845/proving-it-has-all-real-roots/2275912#2275912 – zwim Jul 07 '17 at 22:00

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factorizing the whole Thing we obtaine: $$-ab-ac+2\,ax-bc+2\,bx+2\,xc-3\,{x}^{2}=0$$ can you solve this? solving this equation we get $$x_1=a/3+b/3+c/3+1/3\,\sqrt {{a}^{2}-ab-ac+{b}^{2}-bc+{c}^{2}}$$ or $$x_2=a/3+b/3+c/3-1/3\,\sqrt {{a}^{2}-ab-ac+{b}^{2}-bc+{c}^{2}}$$ note that we have $$a^2+b^2+c^2\geq ab+bc+ca$$

Dr. Sonnhard Graubner
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