In a triangle ABC, $(a+b+c)(b+c-a)=\lambda$  bc if 

We are given that  $(a+b+c)(b+c-a)=\lambda bc,$
 Or  ${(b+c)}^2-a^2=\lambda bc.$ That is , $b^2+c^2+2bc-a^2=\lambda bc,or$  
 $b^2+c^2+-a^2=(\lambda -2)bc$ .
 Therefore,     $\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{\lambda -2}{2}$
 As A is the angle of a triangle, $-1<\cos A<1.$  
 Therefore     $-1<(\lambda -2)/2<1\Rightarrow 0<\lambda <4$