Prove that there do not exist real numbers x y and z such that x plus y plus z equals 0 and simultaneously 1 over x plus 1over y plus 1over z equals 0?

Answer 1

Suppose x + y + z = 0

then z = - x - y = -(x + y) . . . . . . (A)

1/x + 1/y + 1/z = 0

implies x, y and z are all non-zero: otherwise the reciprocals are undefined.

then z ≠ 0 implies that x+y ≠ 0 (by (A))

so 1/x + 1/y + 1/[-(x+y)] = 0 (using (A))

so that 1/x + 1/y = 1/(x+y)

ie (x + y)/xy = 1/(x + y)

Now, since x+y ≠ 0, multiply both sides by x+y to give

(x + y)2/xy = 1

or (x + y)2 = xy

x2 + 2xy + y2 = xy

x2 + xy + y2 = 0

so that x = [-y ± sqrt(y2 - 4y2)]/2

= [-y ± y*sqrt(-3)]/2 which is cannot be real if y is real.