By AM-GM we have:

\(E=\dfrac{4a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{4c}{a+b-c}\)

\(=4\left(\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{4c}{a+b-c}\right)\)

\(\ge4\cdot3\sqrt[3]{\dfrac{abc}{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}\)

\(\ge4\cdot3\sqrt[3]{\dfrac{abc}{abc}}=4\cdot3=12\)

Need prove \(\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)\le abc\)

The last inequality is right follow Schur's inequality

\("="\Leftrightarrow a=b=c\)