Từ \(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
\(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\hept{\begin{cases}x,y,z>0\\xy+yz+xz=1\end{cases}}\)
\(A=\frac{x}{yz+1}+\frac{y}{xz+1}+\frac{z}{xy+1}\)
\(=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\)
\(\ge\frac{\left(x+y+z\right)^2}{3xyz+x+y+z}\)\(\ge\frac{\left(x+y+z\right)^2}{\frac{\left(x+y+z\right)\left(xy+yz+xz\right)}{3}+x+y+z}\)
\(=\frac{3\left(x+y+z\right)}{xy+yz+xz+3}\)\(\ge\frac{3\sqrt{3\left(xy+yz+xz\right)}}{xy+yz+xz+3}\)
\(=\frac{3\sqrt{3}}{1+3}=\frac{3\sqrt{3}}{4}\)