\(a+b+c=2\Rightarrow\left(a+b+c\right)^2=4\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+ac+bc\right)=4\)
\(\Rightarrow ab+ac+bc=\dfrac{4-\left(a^2+b^2+c^2\right)}{2}=\dfrac{4-2}{2}=1\)
\(\Rightarrow\left\{{}\begin{matrix}1+b^2=b^2+ab+ac+bc=\left(a+b\right)\left(b+c\right)\\1+c^2=c^2+ab+ac+bc=\left(a+c\right)\left(b+c\right)\\1+a^2=a^2+ab+ac+bc=\left(a+b\right)\left(a+c\right)\end{matrix}\right.\)
\(\Rightarrow a\sqrt{\dfrac{\left(1+b^2\right)\left(1+c^2\right)}{1+a^2}}=a\sqrt{\dfrac{\left(b+c\right)^2\left(a+b\right)\left(a+c\right)}{\left(a+b\right)\left(a+c\right)}}=a\left(b+c\right)\)
Tương tự ta có: \(b\sqrt{\dfrac{\left(1+a^2\right)\left(1+c^2\right)}{1+b^2}}=b\left(a+c\right)\)
\(c\sqrt{\dfrac{\left(1+a^2\right)\left(1+b^2\right)}{1+c^2}}=c\left(a+b\right)\)
\(\Rightarrow A=a\left(b+c\right)+b\left(a+c\right)+c\left(a+b\right)=2\left(ab+ac+bc\right)=2\)
