# 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 $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 1$ prove that ${a}^{2} / \left(b + c\right) + {b}^{2} / \left(c + a\right) + {c}^{2} / \left(a + b\right) = 0$

Nov 29, 2016

See below.

#### Explanation:

We know that

$\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} - 1 = 0$ which is equivalent to

$\frac{{a}^{3} + {b}^{3} + a b c + {c}^{3}}{\left(a + b\right) \left(a + c\right) \left(b + c\right)} = 0$

also

${a}^{2} / \left(b + c\right) + {b}^{2} / \left(c + a\right) + {c}^{2} / \left(a + b\right) =$
$= \frac{{a}^{4} + {a}^{3} b + a {b}^{3} + {b}^{4} + {a}^{3} c + {a}^{2} b c + a {b}^{2} c + {b}^{3} c + a b {c}^{2} + a {c}^{3} + b {c}^{3} + {c}^{4}}{\left(a + b\right) \left(a + c\right) \left(b + c\right)} = 0$

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

${a}^{2} / \left(b + c\right) + {b}^{2} / \left(c + a\right) + {c}^{2} / \left(a + b\right) = 0$

Nov 29, 2016

Given relation

$\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 1$

Multiplying both sides by $a + b + c$ we get $\left(a + b + c \ne 0\right)$

$\implies \frac{a \left(a + b + c\right)}{b + c} + \frac{b \left(a + b + c\right)}{c + a} + \frac{c \left(a + b + c\right)}{a + b} = \left(a + b + c\right)$

$\implies {a}^{2} / \left(b + c\right) + \frac{a \left(b + c\right)}{b + c} + {b}^{2} / \left(c + a\right) + \frac{b \left(c + a\right)}{c + a} + {c}^{2} / \left(a + b\right) + \frac{c \left(a + b\right)}{a + b} = \left(a + b + c\right)$

$\implies {a}^{2} / \left(b + c\right) + a + {b}^{2} / \left(c + a\right) + b + {c}^{2} / \left(a + b\right) + c = \left(a + b + c\right)$

$\implies {a}^{2} / \left(b + c\right) + {b}^{2} / \left(c + a\right) + {c}^{2} / \left(a + b\right) = 0$

Proved