Đặt \(\left\{{}\begin{matrix}b+c-a=x\\c+a-b=y\\a+b-c=z\end{matrix}\right.\)\(\left(x,y,z>0\right)\)\(\Rightarrow\left\{{}\begin{matrix}x+y=2c\\y+z=2a\\x+z=2b\end{matrix}\right.\)
Thì ta có: \(\dfrac{2\left(y+z\right)}{x}+\dfrac{9\left(x+z\right)}{2y}+\dfrac{8\left(x+y\right)}{z}\ge26\)