Ta có: [tex]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})-3=\frac{1}{2}[(a+b)+(b+c)+(c+a)](\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})-3\geq \frac{1}{2}.9-3=\frac{3}{2}[/tex]
Dấu " = " xảy ra khi a = b = c.
Ta có: [tex]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})-3=\frac{1}{2}[(a+b)+(b+c)+(c+a)](\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})-3\geq \frac{1}{2}.9-3=\frac{3}{2}[/tex]
Dấu " = " xảy ra khi a = b = c.
Vì [tex]\frac{a}{b+c}=\frac{a+b+c}{b+c}-1;\frac{b}{c+a}=\frac{a+b+c}{c+a}-1;\frac{c}{a+b}=\frac{a+b+c}{a+b}-1\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a +b}=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}-3=(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})-3[/tex]