1. **Problem statement:** Two real numbers $a$ and $b$ satisfy the inequality $$(a - b)^2 > a^2 + b^2.$$ We want to determine which of the given statements always holds: - $a = 0$ or $b = 0$ - $a < 0 < b$ or $b < 0 < a$ - $a - b > 1$ - $a - b > 0$ - $b < 0$ 2. **Rewrite the inequality:** $$ (a - b)^2 > a^2 + b^2 $$ Expanding the left side: $$ a^2 - 2ab + b^2 > a^2 + b^2 $$ 3. **Simplify by subtracting $a^2 + b^2$ from both sides:** $$ a^2 - 2ab + b^2 - (a^2 + b^2) > 0 $$ $$ \cancel{a^2} - 2ab + \cancel{b^2} - \cancel{a^2} - \cancel{b^2} > 0 $$ $$ -2ab > 0 $$ 4. **Divide both sides by $-2$ (note the inequality direction changes because dividing by a negative):** $$ ab < 0 $$ 5. **Interpretation:** The product $ab$ is less than zero, which means one of $a$ or $b$ is positive and the other is negative. 6. **Check the given statements:** - $a=0$ or $b=0$: Not necessarily true because $ab<0$ excludes zero. - $a < 0 < b$ or $b < 0 < a$: This matches the condition $ab<0$. - $a - b > 1$: Not necessarily true. - $a - b > 0$: Not necessarily true. - $b < 0$: Not necessarily true because $a$ could be negative instead. **Final answer:** The statement that always holds is: $$ a < 0 < b \quad \text{or} \quad b < 0 < a $$