Ta có: $\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}\iff \frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}=3$
$\iff 1+\frac{c}{a+b}+1+\frac{b}{c+a}=3\iff \frac{c}{a+b}+\frac{b}{a+c}=1$
$\iff \frac{c(a+c)+b(a+b)}{(a+b)(a+c)}=1$
$\iff b^2+c^2=a^2+bc\iff \frac{b^2+c^2-a^2}{2bc}=\frac{1}{2}\iff cos(A)=\frac{1}{2}\iff A=60^0$.
Bài toán được chứng minh hoàn tất.