Đặt:
\(\left\{{}\begin{matrix}b+c-a=x\\a+c-b=y\\a+b-c=z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=2c\\y+z=2a\\x+z=2b\end{matrix}\right.\)
\(A=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
\(2A=\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\)
\(=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{y}{z}+\dfrac{z}{y}\right)\ge2\sqrt{\dfrac{xy}{xy}}+2\sqrt{\dfrac{yz}{yz}}+2\sqrt{\dfrac{xz}{xz}}=6\) (AM-GM)
\(\Rightarrow2A\ge6\Leftrightarrow A\ge3\)
\("="\Leftrightarrow a=b=c\) hay tam giác đã cho là tam giác đều
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