M=$\frac{1}{ab+a+2}$ +$\frac{1}{bc+b+2}$ +$\frac{1}{ca+c+2}$
Ta có:
$\frac{1}{ab+a+2}$ =$\frac{1}{(ab+1)+(a+1)}$ ≤$\frac{1}{4}$($\frac{1}{ab+1}$+$\frac{1}{a+1}$) (BĐT Bunhicopski)
Tương tự:
$\frac{1}{bc+b+2}$ ≤$\frac{1}{4}$($\frac{1}{bc+1}$+$\frac{1}{b+1}$)
$\frac{1}{ca+c+2}$ ≤$\frac{1}{4}$($\frac{1}{ca+1}$+$\frac{1}{c+1}$)
⇒M≤$\frac{1}{4}$($\frac{1}{ab+1}$+$\frac{1}{a+1}$)+$\frac{1}{4}$($\frac{1}{bc+1}$+$\frac{1}{b+1}$)+$\frac{1}{4}$($\frac{1}{ca+1}$+$\frac{1}{c+1}$)=$\frac{1}{4}$($\frac{1}{ab+1}$+$\frac{1}{a+1}$+$\frac{1}{bc+1}$+$\frac{1}{b+1}$+$\frac{1}{ca+1}$+$\frac{1}{c+1}$)=$\frac{1}{4}$($\frac{ab+1}{ab+1}$ +$\frac{bc+1}{bc+1}$ +$\frac{ca+1}{ca+1}$) =$\frac{1}{4}$(1+1+1)=$\frac{3}{4}$
Dấu "=" xảy ra a=b=c=1
Vậy Max M=$\frac{3}{4}$ khi a=b=c=1
