How prove this inequality $(y-z)^2>8(y+z)$ 
For three distinct postive integer $x,y,z$ such
$$(x+y)(x+z)=(y+z)^2$$
show that
$$(y-z)^2>8(y+z)$$

My some idea: since
$$x^2+(y+z)x+yz-(y+z)^2=0$$
so
$$\Delta_{x}=(y+z)^2-4yz+4(y+z)^2=5(y+z)^2-4yz=m^2$$then I can't

WLOG $y<z$, so let $z=y+a$, where $a>0$. Then the discriminant is
\begin{align*}5(y+z)^2-4yz&=5(2y+a)^2-4y(y+a)=5(4y^2+4ya+a^2)-4y^2-4ya\\
&=4(4y^2+4ya+a^2)+a^2=4(2y+a)^2+a^2=(4y+2a)^2+a^2\end{align*}
So if it's a square, then it's either the next square
$$(4y+2a+1)^2=(4y+2a)^2+2(4y+2a)+1$$
and
$$a^2=4(2y+a)+1\implies 8y+5=a^2-4a+4=(a-2)^2,$$
but that's not possible since squares cannot be $5$ mod $8$ (the remainders can only be $0$, $1$, or $4$).
Or it must be at least
$$(4y+2a+2)^2=(4y+2a)^2+4(4y+2a)+4,$$
then
$$(z-y)^2=a^2>4(4y+2a)=8(2y+a)=8(y+z).$$
