\(\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ac}\)

\(=\frac{abc}{abc+a\times abc+ab}+\frac{abc}{abc+b+bc}+\frac{1}{1+c+ac}\)

\(=\frac{abc}{ab\left(c+ac+1\right)}+\frac{abc}{b\left(ac+1+c\right)}+\frac{1}{1+c+ac}\)

\(=\frac{c}{c+ac+1}+\frac{ac}{ac+1+c}+\frac{1}{1+c+ac}\)

\(=\frac{c+ac+1}{c+ac+1}\)

= 1

\(A=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)\(=\frac{1}{ab+a+1}+\frac{a}{a\left(bc+b+1\right)}+\frac{abc}{ca+c+abc}\)

\(=\frac{1}{ab+a+1}+\frac{a}{1+ab+a}+\frac{ab}{a+1+ab}=1\)

Theo bài ra ta có: a.b.c = 1

    =>  a=1;b=1;c=1

Ta có: A = \(\frac{1}{a.b+a+1}\)\(+\frac{1}{b.c+b+1}+\frac{1}{c.a+c+1}\)\(=\frac{1}{1.1+1+1}+\frac{1}{1.1+1+1}\)\(+\frac{1}{1.1+1+1}\)

             \(=\frac{1}{1+1+1}+\frac{1}{1+1+1}+\frac{1}{1+1+1}\)\(=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}=1\)

Vậy A = 1

\(A=\)\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)

    \(=\frac{c}{\left(ab+a+1\right)c}+\frac{ac}{\left(bc+b+1\right).ac}+\frac{1}{ca+c+1}\)

    \(=\frac{c}{abc+ac+c}+\frac{ac}{abc^2+abc+ac}+\frac{1}{ca+c+1}\)

    \(=\frac{c}{1+ac+c}+\frac{ac}{c+1+ac}+\frac{1}{ca+c+1}\)

    \(=\frac{c+ac+1}{1+ac+c}=1\)

Giải:

Từ giả thiết ta có:

\(\left(1-b\right)\left(1-c\right)\ge0\)

\(\Leftrightarrow1-\left(b+c\right)+bc\ge0\)

\(\Leftrightarrow bc+1\ge b+c\)

\(\Rightarrow\frac{a}{bc+1}\le\frac{a}{b+c}\le\frac{a}{a+b}\left(1\right)\)

Tương tự ta có:

\(\frac{b}{ac+1}\le\frac{b}{a+c}\le\frac{b}{a+b}\left(2\right)\)

\(\frac{c}{ab+1}\le c\le1\left(3\right)\)

Cộng theo vế \(\left(1\right);\left(2\right);\left(3\right)\) ta được:

\(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{a+b}{a+b}+1=2\)

Vậy \(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le2\) (Đpcm)