Proving $x^2 - y^2 + z^2 \gt (x - y + z)^2$ Prove that

$$x^2 - y^2 + z^2 > (x - y + z)^2$$

where: $x < y <z$  for all natural numbers.
Thank for help.

I think the formula $A^2 − B^2 = (A− B)(A+ B)$, will be perfect.
$x^2 − y^2 > (x − y + z)^2 − z^2$
$(x − y)(x + y) > (x − y + z − z)(x − y + z + z)$,
$(x − y)(x + y) > (x − y)(x − y + 2z)$.
