có:
[TEX](a+b+c+3)^2[/TEX]
[TEX]=a^2+b^2+c^2+9+2ab+2bc+2ca+6a+6b+6c[/TEX]
[TEX]=12+2ab+2bc+2ca+6a+6b+6c[/TEX]
[TEX]=2(6+ab+bc+ca+3a+3b+3c)[/TEX]
[TEX]=2((a+b+1)(b+1)+(b+c+1)(c+1)+(c+a+1)(a+1))[/TEX]
[TEX]\Rightarrow \frac{(a+b+c+3)^2}{(a+b+1)(b+1)+(b+c+1)(c+1)+(c+a+1)(a+1)}=2[/TEX]
[TEX]\Rightarrow 2 \leq \frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}[/TEX]
[TEX]\Leftrightarrow 1 \geq \frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}[/TEX]
[TEX]\Leftrightarrow \frac{1}{2} \geq \frac{a}{2a+2b+21}+\frac{b}{2b+2c+2}+\frac{c}{2c+2a+2}[/TEX]
[TEX]\Leftrightarrow \frac{1}{2} \geq \frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{c}{c^2+2a+3}[/TEX]