a, b are positive reals. Show that 1/x + 1/(x-a) + 1/(x+b) = 0 has two real roots one in [a/3, 2a/3] and the other in [-2b/3, -b/3].
Solution
Multiplying up the equation is 3x2 + 2(b-a)x - ab = 0 with roots (a-b ±√(a2+ab+b2))/3. Note that these are both real since a and b are both positive. Consider first the larger root k. We have a2+ab+b2 < a2+2ab+b2, so k < (a-b+a+b)/3 = 2a/3. Also a2+ab+b2 > b2, so k > (a-b+b)/3 = a/3. Hence k ∈ (a/3, 2a/3).
Let h be the smaller root. Then h > (a-b-(a+b))/3 = -2b/3 and since a2+ab+b2 > a2, h < (a-b-a)/3 = -b/3, so h ∈ (-2b/3, -b/3).
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03