Question

If `a/(b+c) + b/(c+a) + c/(a+b)=1` prove that `a^2/(b+c) + b^2/(c+a) + c^2/(a+b)=0` ?

If `a/(b+c) + b/(c+a) + c/(a+b)=1` prove that
`a^2/(b+c) + b^2/(c+a) + c^2/(a+b)=0`

Answer 1

See below.

Explanation:

We know that

`a/(b+c) + b/(c+a) + c/(a+b)-1=0` which is equivalent to

`(a^3 + b^3 + a b c + c^3)/((a + b) (a + c) (b + c))=0`

also

`a^2/(b+c) + b^2/(c+a) + c^2/(a+b)=`
`=(a^4 + a^3 b + a b^3 + b^4 + a^3 c + a^2 b c + a b^2 c + b^3 c + a b c^2 + a c^3 + b c^3 + c^4)/((a + b) (a + c) (b + c))=0`

but

`a^4 + a^3 b + a b^3 + b^4 + a^3 c + a^2 b c + a b^2 c + b^3 c + a b c^2 + a c^3 + b c^3 + c^4=(a+b+c)(a^3 + b^3 + a b c + c^3)`

so

`a^4 + a^3 b + a b^3 + b^4 + a^3 c + a^2 b c + a b^2 c + b^3 c + a b c^2 + a c^3 + b c^3 + c^4=0`

and consequently

`a^2/(b+c) + b^2/(c+a) + c^2/(a+b)=0`

Answer 2

Given relation

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

Multiplying both sides by `a+b+c` we get `(a+b+c!=0)`

`=>(a(a+b+c))/(b+c)+(b(a+b+c))/(c+a)+(c(a+b+c))/(a+b)=(a+b+c)`

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

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

`=>a^2/(b+c)+b^2/(c+a)+c^2/(a+b)=0`

Proved