SOLUTION: prove that if a,b,c are real, the roots of (1/x+a) +(1/x+b) +(1/x+c) =3/x are also real
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Question 464215: prove that if a,b,c are real, the roots of (1/x+a) +(1/x+b) +(1/x+c) =3/x are also real
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
I presume you mean
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:
 + (\frac{1}{x+b} - \frac{1}{x}) + (\frac{1}{x+c} - \frac{1}{x}) = 0)
} + \frac{b}{x(x+b)} + \frac{c}{x(x+c)} = 0)
(combined fractions, then multiplied by -1)
Provided 
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):
(x+c) + b(x+a)(x+c) + c(x+a)(x+b) = 0)
This will result in a nice second-degree polynomial:
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,
^2 - 12(a+b+c)abc \ge 0 )
^2 - 3(a+b+c)abc \ge 0)
)
Note that this is true, since we can apply the Cauchy-Schwarz inequality, which tells us that
^2)
(shorthand for cyclic sum)
)
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.
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