Prove that $\alpha$ lies between $0$ and $4$. 
Let $a,b,c$ be the length of the sides of the triangle $ABC$ .
  Given $(a+b+c)(b+c-a)=\alpha bc$.Then Prove that the value of $\alpha$ lies in between $0$ and $4$.


$\begin{align}(a+b+c)(b+c-a)&=\alpha bc\\
\implies \alpha&=\dfrac{b^2+c^2-a^2}{bc}+2\\
 \alpha&=2\cos A+2\\
\end{align}$
I also know a relation like $a+b>c\\b+c>a\\a+c>b$
I have studied maths up to $12th$ grade.

As given $(a+b+c)(b+c-a)=\alpha bc$ then we can simply as follows 
$$(b+c)^2-a^2=\alpha bc \implies \alpha=\frac{b^2+c^2+2bc-a^2}{bc}=\frac{b^2+c^2-a^2}{bc}+2=2+2\cos A$$ Now, for a triangle to exist, we have a condition for angle $A$ as $0<A<\pi$ thus we get $-1<\cos A<1$ hence, we get 
$$(2-2)<\alpha<(2+2) \implies 0<\alpha<4$$ Thus, the value of $\alpha$ lies between 0 & 4.
But including two extreme values for a triangle being a line, we get $0\leq\alpha\leq 4$ which shows that the triangle is a line for two extreme values $\alpha=0$ & $\alpha=4$.   
