Croatian Mathematical Competitions

We need to prove that
$$c^2 - b^2 + a^2 \ge (c - b + a)^2.$$

Expand the right-hand side:
$$(c - b + a)^2 = (c + a - b)^2 = (c + a)^2 - 2b(c + a) + b^2 = c^2 + 2ac + a^2 - 2bc - 2ab + b^2.$$

So the inequality becomes:
$$c^2 - b^2 + a^2 \ge c^2 + 2ac + a^2 - 2bc - 2ab + b^2.$$

Bring all terms to one side:
$$c^2 - b^2 + a^2 - [c^2 + 2ac + a^2 - 2bc - 2ab + b^2] \ge 0.$$

Simplify:
$$c^2 - b^2 + a^2 - c^2 - 2ac - a^2 + 2bc + 2ab - b^2 \ge 0$$
$$(- b^2 - 2ac + 2bc + 2ab - b^2) \ge 0$$
$$(-2b^2 - 2ac + 2bc + 2ab) \ge 0$$
$$2(-b^2 - ac + bc + ab) \ge 0$$
$$-b^2 - ac + bc + ab \ge 0$$
$$(bc + ab - b^2 - ac) \ge 0$$
$$b(c + a - b) - ac \ge 0$$
$$b(c + a - b) \ge ac$$

Since $a \le b \le c$, $c + a - b \ge a$ (because $c \ge b$), and $b \ge a$, so $b(c + a - b) \ge a b$. Also, $ac \le b c$ since $a \le b$. So $b(c + a - b) \ge ac$.

Therefore, the inequality holds for all real numbers $a \le b \le c$.