Question

if 1/(a+1)+1/(b+1)+1/(c+1)=2 then a^2+b^2+c^2

Answer 1

Answer:

Write u = 1/a, v = 1/b, w = 1/c. Then 1/(a+b+c) = 1/(1/u + 1/v + 1/w) = uvw/(vw + uw + uv).

Step-by-step explanation:

1/a+1/b+1/c=1/(a+b+c)

or (a+b+c)(ab+bc+ca)

from this equation I can write, (a+b)^2+(b+c)^2+(c+a)^2=0