Có $\frac 12=\cos B=\frac{a^2+c^2-b^2}{2ac}$$\Rightarrow ac=a^2+c^2-b^2$
$\Rightarrow 3ac=(a+c)^2-b^2=(a+b+c)(a+c-b)$
$\Rightarrow \frac 3{a+b+c}=\frac{a+c-b}{ac}=\frac 1a +\frac 1c -\frac{b}{ac}(*)$
Ta lại có $\frac 1{ac}=\frac 1{a(a+b)}+\frac 1{c(b+c)}\Leftrightarrow a^2+c^2-b^2=ac$(luôn đúng)
Do đó $(*)\Leftrightarrow \frac{3}{a+b+c}=[\frac 1a -\frac{b}{a(a+b)}]+[\frac 1c-\frac b{c(b+c)}]=\frac 1{a+b}+\frac 1{b+c}$ (đpcm)