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cho [TEX]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1[/TEX]
chứng minh:

[TEX]\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0[/TEX]

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

$\leftrightarrow (a + b + c)(\dfrac{a}{b + c} + \dfrac{b}{a + c} + \dfrac{c}{a + b}) = a + b + c$

$\leftrightarrow \dfrac{a^2 + a(b + c)}{b + c} + \dfrac{b^2 + b(a + c)}{a + c} + \dfrac{c^2 + c(a + b)}{a + b} = a + b + c$

$\leftrightarrow \dfrac{a^2}{b + c} + a + \dfrac{b^2}{a + c} + b + \dfrac{c^2}{a + b} + c = a + b + c$

$\leftrightarrow \dfrac{a^2}{b + c} + \dfrac{b^2}{a + c} + \dfrac{c^2}{a + b} = 0$
 
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