Question

Prove that :if (x+y)(y+z)(x+z)=0 x,y,z equal to not zero then 1/x+1/y+1/z=1/(x+y+z)

Answer 1

@marcelie

Answer 2

@zepdrix

Answer 3

you can do it by cases.
if (x+y)(y+z)(x+z)=0 and x,y,z \( \ne\) 0
then we have 3 cases:
x+y=0
y+z=0
x+z=0

Answer 4

or any combination of the three terms may be zero

assume x+y=0 so x= -y
then
\[ \frac{1}{x}+\frac{1}{y}+\frac{1}{z} = - \frac{1}{y}+\frac{1}{y}+\frac{1}{z} \\
= \frac{1}{z} \]
and because x+y=0 we can say
\[ \frac{1}{z} = \frac{1}{x+y+z} \]
and thus
\[ \frac{1}{x}+\frac{1}{y}+\frac{1}{z} = \frac{1}{x+y+z} \]

the other two cases give the same result (you could argue "by symmetry")

to finish, show that if both (x+y) and (y+z)=0 you get the same result, (and similarly for any pair)

and finally, if all 3 terms = 0

Answer 5

though maybe we don't have to consider "multiple zeros"
as the first proof should also cover those case.