Xin lỗi bạn , mik giải muộn 
Do a ; b ; c là 3 cạnh của tam giác \(\Rightarrow a+b-c>0;b+c-a>0;c+a-b>0\)
Đặt \(a+b-c=x;b+c-a=y;a+c-b=z\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+z=a+b+c\\x+y=2b;y+z=2c;z+x=2a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+z=a+b+c\\b=\dfrac{x+y}{2};c=\dfrac{y+z}{2};a=\dfrac{x+z}{2}\end{matrix}\right.\)
Đặt PT đã cho là A . Ta có :
\(A=\dfrac{\left(x+y\right)\left(x+z\right)}{4x}+\dfrac{\left(x+y\right)\left(y+z\right)}{4y}+\dfrac{\left(y+z\right)\left(x+z\right)}{4z}\)
\(=\dfrac{x^2+xy+xz+yz}{4x}+\dfrac{xy+y^2+xz+yz}{4y}+\dfrac{xy+xz+yz+z^2}{4z}\)
\(=\dfrac{x+y+z}{4}+\dfrac{yz}{4x}+\dfrac{x+y+z}{4}+\dfrac{xz}{4y}+\dfrac{x+y+z}{4}+\dfrac{xy}{4z}\)
\(=\dfrac{3}{4}\left(x+y+z\right)+\dfrac{yz}{4x}+\dfrac{xz}{4y}+\dfrac{xy}{4z}\)
\(=3\left(x+y+z\right)+\dfrac{y^2z^2+x^2z^2+x^2y^2}{4xyz}\)
Áp dụng BĐT phụ \(a^2+b^2+c^2\ge ab+bc+ac\)
\(\Rightarrow x^2y^2+y^2z^2+x^2z^2\ge xyz\left(x+y+z\right)\)
\(\Rightarrow A\ge\dfrac{3}{4}\left(x+y+z\right)+\dfrac{x+y+z}{4}=x+y+z=a+b+c\)
\(\Rightarrowđpcm\)
