Ta có \(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=2\Leftrightarrow\dfrac{1}{a+1}=1-\dfrac{1}{b+1}+1-\dfrac{1}{c+1}\Leftrightarrow\dfrac{1}{a+1}=\dfrac{b+1-1}{b+1}+\dfrac{c+1-1}{c+1}\Leftrightarrow\dfrac{1}{a+1}=\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge2\sqrt{\dfrac{b.c}{\left(b+1\right)\left(c+1\right)}}\)(bđt cosi)
chứng minh tương tự: \(\dfrac{1}{b+1}\ge2\sqrt{\dfrac{ac}{\left(a+1\right)\left(c+1\right)}}\)
\(\dfrac{1}{c+1}\ge2\sqrt{\dfrac{ab}{\left(a+1\right)\left(b+1\right)}}\)
Vậy \(\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge8\sqrt{\dfrac{b.c.c.a.a.b}{\left(c+1\right)\left(b+1\right)\left(a+1\right)\left(c+1\right)\left(a+1\right)\left(b+1\right)}}=\dfrac{8.abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\Leftrightarrow1\ge8abc\Rightarrow abc\le\dfrac{1}{8}\Rightarrow P\le\dfrac{1}{8}\)
dấu = xảy ra khi \(\dfrac{a}{a+1}=\dfrac{b}{b+1}=\dfrac{c}{c+1},abc=\dfrac{1}{8}\Leftrightarrow a=b=c=\dfrac{1}{2}\)
Vậy gtln của P=abc là \(\dfrac{1}{8}\)
