M=1 khi và chỉ khi abc=1
Áp dụng giả thiết từ đề bài :
\(M=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(\Leftrightarrow M=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}\)
\(\Leftrightarrow M=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+b}\)
\(\Leftrightarrow M=\frac{1+b+bc}{b+1+bc}=1\)
Vậy M = 1
Ta có : \(M=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)
\(=\frac{1}{ab+a+1}+\frac{1}{\frac{1}{a}+b+1}+\frac{1}{c+ca+a.b.c}\)
\(=\frac{1}{ab+a+1}+\frac{a}{ab+a+1}+\frac{1}{c.\left(ab+a+1\right)}\)
\(=\frac{1}{ab+a+1}+\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}\)
\(=\frac{a+ab+1}{a+ab+1}=1\)
Vậy M = 1
