Đặt \(\left(a+1;b+1;c+1\right)\rightarrow\left(x;y;z\right)\).Giả thiết trở thành:\(xyz=x+y+z\) và cần tìm max của \(P=\sum\dfrac{x}{x^2+1}\)
Ta có: \(P=\sum\dfrac{x}{x^2+1}=\sum\dfrac{xyz}{x\left(x+y+z\right)+yz}=xyz.\sum\dfrac{1}{\left(x+y\right)\left(x+z\right)}\)
\(=\dfrac{2xyz\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
Do \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\) nên \(P\le\dfrac{2xyz}{\dfrac{8}{9}\left(xy+yz+xz\right)}=\dfrac{9}{4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}\)(*)
Mặt khác , từ giả thiết ta có : \(1=\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\le\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\)( theo AM-GM)
\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\sqrt{3}\)
Kết hợp với (*) , ta suy ra \(P\le\dfrac{9}{4\sqrt{3}}=\dfrac{3\sqrt{3}}{4}\)
Dấu = xảy ra khi \(x=y=z=\sqrt{3}\) hay \(a=b=c=\sqrt{3}-1\)
P/s: Chứng minh \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
khai triển ra ta có: \(\sum ab\left(a+b\right)\ge6abc\)hay \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)( đúng)
