Ta có: [tex]\frac{a}{a+b}> \frac{a}{a+b+c};\frac{b}{b+c}> \frac{b}{a+b+c};\frac{c}{c+a}> \frac{c}{a+b+c}\Rightarrow \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a+b+c}{a+b+c}=1[/tex]
Cần chứng minh BĐT phụ sau: [tex]\frac{x}{y}< \frac{x+z}{y+z}[/tex] với x < y.
Thật vậy, [tex]\frac{x+z}{y+z}-\frac{x}{y}=\frac{y(x+z)-x(y+z)}{y(y+z)}=\frac{z(y-z)}{y(y+z)}> 0\Rightarrow \frac{x+z}{y+z}> \frac{x}{y}[/tex]
Áp dụng vào bài ta có:[tex]\frac{a}{a+b}< \frac{a+c}{a+b+c},\frac{b}{b+c}< \frac{a+b}{a+b+c},\frac{c}{c+a}< \frac{b+c}{a+b+c}\Rightarrow \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+b+b+c+c+a}{a+b+c}=2[/tex]