If $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 1$ $a/(b+c) + b/(c+a) + c/(a+b)=1$ prove that $\frac{a^{2}}{b + c} + \frac{b^{2}}{c + a} + \frac{c^{2}}{a + b} = 0$ $a^2/(b+c) + b^2/(c+a) + c^2/(a+b)=0$ ?

Question

If $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 1$ $a/(b+c) + b/(c+a) + c/(a+b)=1$ prove that $\frac{a^{2}}{b + c} + \frac{b^{2}}{c + a} + \frac{c^{2}}{a + b} = 0$ $a^2/(b+c) + b^2/(c+a) + c^2/(a+b)=0$ ?

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

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Answer 1 · 2016-11-29T16:17:41

See below.

Explanation:

We know that

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

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

also

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

but

$a^{4} + a^{3} b + a b^{3} + b^{4} + a^{3} c + a^{2} b c + a b^{2} c + b^{3} c + a b c^{2} + a c^{3} + b c^{3} + c^{4} = (a + b + c) (a^{3} + b^{3} + a b c + c^{3})$ $a^4 + a^3 b + a b^3 + b^4 + a^3 c + a^2 b c + a b^2 c + b^3 c + a b c^2 + a c^3 + b c^3 + c^4=(a+b+c)(a^3 + b^3 + a b c + c^3)$

so

$a^{4} + a^{3} b + a b^{3} + b^{4} + a^{3} c + a^{2} b c + a b^{2} c + b^{3} c + a b c^{2} + a c^{3} + b c^{3} + c^{4} = 0$ $a^4 + a^3 b + a b^3 + b^4 + a^3 c + a^2 b c + a b^2 c + b^3 c + a b c^2 + a c^3 + b c^3 + c^4=0$

and consequently

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

Answer 2 · 2016-11-29T16:48:34

Given relation

$\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 1$ $a/(b+c)+b/(c+a)+c/(a+b)=1$

Multiplying both sides by $a + b + c$ $a+b+c$ we get $(a + b + c \neq 0)$ $(a+b+c!=0)$

$\Rightarrow \frac{a (a + b + c)}{b + c} + \frac{b (a + b + c)}{c + a} + \frac{c (a + b + c)}{a + b} = (a + b + c)$ $=>(a(a+b+c))/(b+c)+(b(a+b+c))/(c+a)+(c(a+b+c))/(a+b)=(a+b+c)$

$\Rightarrow \frac{a^{2}}{b + c} + \frac{a (b + c)}{b + c} + \frac{b^{2}}{c + a} + \frac{b (c + a)}{c + a} + \frac{c^{2}}{a + b} + \frac{c (a + b)}{a + b} = (a + b + c)$ $=>a^2/(b+c)+(a(b+c))/(b+c)+b^2/(c+a)+(b(c+a))/(c+a)+c^2/(a+b)+(c(a+b))/(a+b)=(a+b+c)$

$\Rightarrow \frac{a^{2}}{b + c} + a + \frac{b^{2}}{c + a} + b + \frac{c^{2}}{a + b} + c = (a + b + c)$ $=>a^2/(b+c)+a+b^2/(c+a)+b+c^2/(a+b)+c=(a+b+c)$

$\Rightarrow \frac{a^{2}}{b + c} + \frac{b^{2}}{c + a} + \frac{c^{2}}{a + b} = 0$ $=>a^2/(b+c)+b^2/(c+a)+c^2/(a+b)=0$

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If $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 1$ $a/(b+c) + b/(c+a) + c/(a+b)=1$ prove that $\frac{a^{2}}{b + c} + \frac{b^{2}}{c + a} + \frac{c^{2}}{a + b} = 0$ $a^2/(b+c) + b^2/(c+a) + c^2/(a+b)=0$ ?