If distinct numbers $a,b,c\in\mathbb N^+$ satisfy $(a+b)(a+c)=(b+c)^2$, prove that $(b-c)^2>8(b+c)$. 
If distinct numbers $a,b,c\in\mathbb N^+$ satisfy $$(a+b)(a+c)=(b+c)^2$$prove that $$(b-c)^2>8(b+c).$$

The first thing I did after I saw the problem was turning the inequality into this: $$(b+c)^2-8(b+c)-4bc>0$$
After a lot of different approaches I still can't find a way to solve this.

$(a+b)(a+c)=(b+c)^2\implies a^2 +a(b+c)+bc-(b+c)^2=0$, which is a quadratic in $a$. Solving for $a=\frac {-2(b+c)+\sqrt{(b+c)^2-4bc+4(b+c)^2}}{2}=\frac {-2(b+c)+\sqrt{(b-c)^2+4(b+c)^2}}{2}$, (we take the positive root since $a\ge1$).
or,  $\frac {-2(b+c)+\sqrt{(b+c)^2-4bc+4(b+c)^2}}{2}\ge1$
or,  $(b-c)^2+4(b+c)^2\ge(2+2b+2c)^2$
or, $(b-c)^2 \ge 4(1+b+c)^2-4(b+c)^2=4(1+2b+2c)>8(b+c).$
