Question

If `a^2/(b+c)=b^2/(c+a)=c^2/(a+b)` then show that `1/(1+a)+1/(1+b)+1/(1+c)=1` ?

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

Answer 1

See below.

Explanation:

From

`a^2/(b+c)=b^2/(c+a)=c^2/(a+b)`

we conclude

`(b^2-a^2)/(a-b) = (c^2-a^2)/(a-c) =(c^2-b^2)/(b-c)`

hence `a=b=c = zeta`

then

`3/(1+zeta) = 1 rArr zeta = 2`

Concluding. The assertion is true only if `a=b=c=2`

NOTE

`u/v = p/q=lambda rArr (upmp)/(vpmq)=lambda`

because

`(upmp)/(vpmq) = (lambda vpmlambda q)/(vpmq) = lambda`

then

`{a^2/(b+c)=b^2/(c+a)=c^2/(a+b)} rArr {(b^2-a^2)/(a-b) = (c^2-a^2)/(a-c) =(c^2-b^2)/(b-c)}`

but

`{(b^2-a^2)/(a-b) = (c^2-a^2)/(a-c) =(c^2-b^2)/(b-c)} rArr {b+a=c+a=c+b} rArr {a=b=c}`