Put \(A=\dfrac{ab}{a+b-c}+\dfrac{bc}{b+c-a}+\dfrac{ac}{c+a-b}\)
Because a ; b ; c are the length of a triangle so \(a+b-c;b+c-a;c+a-b>0\)
Put \(a+b-c=x;b+c-a=y;c+a-b=z\)
\(\Rightarrow\dfrac{x+y}{2}=b;\dfrac{y+z}{2}=c;\dfrac{x+z}{2}=a;a+b+c=x+y+z\)
We have : \(\dfrac{\left(x+y\right)\left(x+z\right)}{2.2x}+\dfrac{\left(x+y\right)\left(y+z\right)}{2.2y}+\dfrac{\left(x+z\right)\left(y+z\right)}{2.2z}\)
\(=\dfrac{x\left(x+y+z\right)+yz}{4x}+\dfrac{y\left(x+y+z\right)+xz}{4y}+\dfrac{z\left(x+y+z\right)+xy}{4z}\)
\(=\dfrac{x+y+z}{4}+\dfrac{x+y+z}{4}+\dfrac{x+y+z}{4}+\dfrac{yz}{4x}+\dfrac{xz}{4y}+\dfrac{xy}{4z}\)
\(=\dfrac{3\left(x+y+z\right)}{4}+\dfrac{y^2z^2}{4xyz}+\dfrac{x^2z^2}{4xyz}+\dfrac{x^2y^2}{4xyz}\)
Applying the inequality \(a^2+b^2+c^2\ge ab+bc+ac\) , we have :
\(A\ge\dfrac{3\left(x+y+z\right)}{4}+\dfrac{xyz\left(x+y+z\right)}{4xyz}=x+y+z=a+b+c\)
Equal sign occurs \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)
We have :