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

已知 $a$、$b$、$c$ 三數為實數，且
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$，求

$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$ 之值。

【解】

$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$$ $$\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1=1+3$$ $$\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}=4$$ $$\Rightarrow (a+b+c)\left (\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right )=4$$

令 $x=a+b+c$，則

$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$$ $$=\frac{[(x-(b+c)]^2}{b+c}+\frac{[x-(c+a)]^2}{c+a}+\frac{[x-(a+b)]^2}{a+b}$$ $$=\frac{x^2-2(b+c)x+(b+c)^2}{b+c}+\frac{x^2-2(c+a)x+(b+c)^2}{c+a}+\frac{x^2-2(a+b)x+(a+b)^2}{a+b}$$ $$=x^2\left (\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-2x-2x-2x+(b+c+c+a+a+b)$$ $$=x(a+b+c)\left (\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right )-6x+2(a+b+c)$$ $$=4x-6x+2x$$ $$=0$$