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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 $b(c+a-b)\ge ab$. Also, $ac\le bc$ since $a\le b$. So $b(c+a-b)\ge ac$.

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