\(a^2-b^2-c^2+2bc\)
\(=a^2-\left(b^2-2bc+c^2\right)\)
\(=a^2-\left(b-c\right)^2\)
\(=\left(a-b+c\right)\left(a+b-c\right)\)
Vì a, b, c là 3 cạnh của tam giác nên :
\(\left\{{}\begin{matrix}a+c>b\\a+b>c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b+c>0\\a+b-c>0\end{matrix}\right.\)
\(\Rightarrow\left(a-b+c\right)\left(a+b-c\right)>0\)
\(\Leftrightarrow a^2-b^2-c^2+2bc>0\)(đpcm)