\(ab+bc+ca=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
\(P=\dfrac{yz}{x+1}+\dfrac{zx}{y+1}+\dfrac{xy}{z+1}=\dfrac{yz}{x+y+x+z}+\dfrac{zx}{x+y+y+z}+\dfrac{xy}{x+y+y+z}\)
\(P\le\dfrac{1}{4}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)+\dfrac{1}{4}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)
\(P\le\dfrac{1}{4}\left(\dfrac{yz+zx}{x+y}+\dfrac{yz+xy}{x+z}+\dfrac{xz+xy}{y+z}\right)=\dfrac{1}{4}\left(x+y+z\right)=\dfrac{1}{4}\)
\(P_{max}=\dfrac{1}{4}\) khi \(x=y=z=\dfrac{1}{3}\) hay \(x=y=z=3\)
