<s>[INST] 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.
 [/INST] We have
$$(b+c)^2-a^2 = \alpha bc \implies b^2+c^2-a^2 = bc(\alpha-2) \implies \alpha = 2 + \dfrac{b^2+c^2-a^2}{bc} = 2+2\cos(A)$$
Note that $\cos(A) \in [-1,1]$. Hence, we have
$$\alpha \in \left[2-2,2+2\right] = \left[0,4\right]$$
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