c2:
Ta có [tex]\frac{a^{2}}{b+c}=a.\frac{a}{b+c}=a(\frac{a}{b+c}+1-1)=a(\frac{a+b+c}{b+c}-1)[/tex]
=[tex](a+b+c).\frac{a}{b+c}-a[/tex]
Tương tự [tex]\frac{b^{2}}{a+c}=(a+b+c).\frac{b}{a+c}-b[/tex]
[tex](a+b+c).\frac{c}{a+b}-c[/tex]
Suy ra [tex]\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}=(a+b+c)(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})-(a+b+c)[/tex]
=[tex](a+b+c)-(a+b+c)=0[/tex]