Question

If \(\Large \frac{a}{b+c} +\frac{b}{a+c}+\frac{c}{a+b}=1\), then fine
\(\Large \frac{a^2}{b+c} +\frac{b^2}{a+c}+\frac{c^2}{a+b}\)

Answer 1

Gotcha! :D

Answer 2

Let \(\large 1=\frac{a+b+c}{a+b+c} \)

Take \(\large a+b+c\) of the denominotor to the LHS and multiply with every term:
\(\Large \frac{a(a+b+c)}{b+c} +\frac{b(a+b+c)}{a+c} + \frac{c(a+b+c)}{a+b}=a+b+c\)

Then split the terms according to the requirements of the question and you will get:
\(\Large \frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+\frac{a(b+c)}{b+c}+\frac{b(a+c)}{a+c}+\frac{c(a+b)}{a+b}=a+b+c\\\large \Rightarrow \frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=a+b+c-(a+b+c)=0\)

Answer 3

How did you come up with the idea to let 1=(a+b+c)/(a+b+c) to be useful to you?

Answer 4

At any rate, this is awesome, thanks haha.