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#

2 Answers
Nov 29, 2016

Answer:

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#

Nov 29, 2016

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