Question 464215
I presume you mean

*[tex \LARGE \frac{1}{x+a} + \frac{1}{x+b} + \frac{1}{x+c} = \frac{3}{x}]


in which we want to prove that all roots x that satisfy are real. Suppose we rewrite 3/x as 1/x + 1/x + 1/x:


*[tex \LARGE \frac{1}{x+a} + \frac{1}{x+b} + \frac{1}{x+c} = \frac{1}{x} + \frac{1}{x} + \frac{1}{x}]


*[tex \LARGE (\frac{1}{x+a} - \frac{1}{x}) + (\frac{1}{x+b} - \frac{1}{x}) + (\frac{1}{x+c} - \frac{1}{x}) = 0]


*[tex \LARGE \frac{a}{x(x+a)} + \frac{b}{x(x+b)} + \frac{c}{x(x+c)} = 0] (combined fractions, then multiplied by -1)


Provided *[tex x \not= 0] we can multiply both sides by x to cancel the x terms out. After this, we can multiply both sides by (x+a)(x+b)(x+c):


*[tex \LARGE a(x+b)(x+c) + b(x+a)(x+c) + c(x+a)(x+b) = 0]


This will result in a nice second-degree polynomial:


*[tex \LARGE (a+b+c)x^2 + 2(ab + bc + ca)x + 3abc = 0]



All that is left to do is prove that the discriminant of this quadratic is nonnegative, in other words,


*[tex \LARGE 4(ab + bc + ca)^2 - 12(a+b+c)abc \ge 0 ]


*[tex \LARGE (ab + bc + ca)^2 - 3(a+b+c)abc \ge 0]


*[tex \LARGE a^2b^2 + b^2c^2 + c^2a^2 - a^2bc - ab^2c - abc^2 \ge 0]


*[tex \LARGE a^2b^2 + b^2c^2 + c^2a^2 \ge a^2bc + ab^2c + abc^2 (*)]


Note that this is true, since we can apply the Cauchy-Schwarz inequality, which tells us that


*[tex \LARGE \sum_{a,b,c} a^2b^2\sum_{a,b,c} b^2c^2 \ge (\sum_{a,b,c} ab^2c)^2] (shorthand for cyclic sum)


*[tex \LARGE a^2b^2 + b^2c^2 + c^2a^2 \ge ab^2c + abc^2 + a^2bc (*)]


Hence, this implies that the discriminant is positive and the roots of the quadratic are both real. However I do not quite remember if the Cauchy-Schwarz inequality can apply for negative a,b,c (I'm pretty sure it does though, unlike AM-GM), but if it doesn't you might be able to generalize for negative a,b,c.