Vì [tex]a,b,c\geq 1\Rightarrow abc\geq 1[/tex]
Ta có: [tex]T=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\\=\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{a^2bc+abc+ab}[/tex]
Vì [tex]abc\geq 1\Rightarrow abc+ab+a\geq 1+ab+a\Rightarrow \frac{a}{abc+ab+a}\leq \frac{a}{1+ab+a}[/tex]
[tex]abc\geq 1\Rightarrow a^2bc+abc+ab\geq a+1+ab\Rightarrow \frac{ab}{a^2bc+abc+ab}\leq \frac{ab}{a+1+ab}[/tex]
Nên [TEX]T\leq \frac{1}{ab+a+1}+\frac{a}{1+ab+a}+\frac{ab}{a+1+ab}=1[/TEX]
Dấu = xảy ra khi [tex]a=b=c=1[/tex]