Ta có:
$\dfrac{b^2+c^2-a^2}{2bc}+\dfrac{c^2+a^2-b^2}{2ac}+\dfrac{a^2+b^2-c^2}{2ab}=1$
$\Leftrightarrow \dfrac{b^2+c^2-2bc-a^2}{2bc}+\dfrac{c^2+a^2-2ca-b^2}{2ac}+\dfrac{a^2+b^2+2ab-c^2}{2ab}=0$
$\Leftrightarrow \dfrac{(b-c)^2-a^2}{2bc}+\dfrac{(c-a)^2-b^2}{2ac}+\dfrac{(a+b)^2-c^2}{2ab}=0$
$\Leftrightarrow \dfrac{(b-c-a)(b-c+a)}{2bc}+\dfrac{(c-a-b)(c-a+b)}{2ac}+\dfrac{(a+b-c)(a+b+c)}{2ab}=0$
$\Leftrightarrow \dfrac{(a+b-c)[a(b-c-a)-b(c-a+b)+c(a+b+c)]}{2abc}=0$
$\Leftrightarrow \dfrac{(a+b-c)[a(b-c-a)-b(-c-a+b)+c(a-b+c)]}{2abc}=0$
$\Leftrightarrow \dfrac{(a+b-c)(a-b+c)(b+c-a)}{2abc}=0$
$\Leftrightarrow \left[\begin{array}{l}a+b=c\\a+c=b\\b+c=a\end{array}\right.$
Với $a+b=c$, ta có:
$\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{b^2+c^2-(b+c)^2}{2bc}=\dfrac{-2bc}{2bc}=-1$
$\dfrac{c^2+a^2-b^2}{2ac}=\dfrac{c^2+(b+c)^2-b^2}{2(b+c)c}=\dfrac{2c^2+2bc}{2(b+c)c}=1$
$\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{(b+c)^2+b^2-c^2}{2(b+c)c}=\dfrac{2b^2+2bc}{2(b+c)c}=1$
Với $a+c=b$ hoặc $b+c=a$, tương tự.