We are tasked with determining which statement always holds true if $a$ and $b$ satisfy the inequality:

$(a + b)^2 < a^2 + b^2$

Step 1: Expand the inequality

Expand $(a + b)^2$: $(a + b)^2 = a^2 + 2ab + b^2$

The inequality becomes: $a^2 + 2ab + b^2 < a^2 + b^2$

Cancel $a^2 + b^2$ from both sides: $2ab < 0$

Step 2: Analyze the inequality

The condition $2ab < 0$ implies: $ab < 0$

This means $a$ and $b$ must have opposite signs (one is positive and the other is negative).

Step 3: Evaluate each statement

1. $a + b < 0$:
   This does not always hold. For example, if $a = 2$ and $b = -1$, $a + b = 1$, which is not less than 0.

2. $a < 0 \text{ and } b < 0$:
   This cannot hold because $ab < 0$ requires $a$ and $b$ to have opposite signs.

3. $a = 0 \text{ or } b = 0$:
   This does not always hold. For example, $a = 1$ and $b = -1$ satisfy $ab < 0$, but neither $a$ nor $b$ is zero.

4. $a < 0 < b \text{ or } b < 0 < a$:
   This always holds. Since $ab < 0$, $a$ and $b$ must have opposite signs, which means one is negative and the other is positive.

5. $a + b < 1$:
   This does not always hold. For example, $a = 2$ and $b = -1$ satisfy $ab < 0$, but $a + b = 1$, not less than 1.

Final Answer:

The statement that always holds is:

$\boxed{a < 0 < b \text{ or } b < 0 < a}$

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Would you like me to provide a detailed graph or example to further illustrate this?

Here are five related questions to explore:

1. What happens if $a + b = 0$? Does the inequality still hold?
2. Can you derive the inequality $ab < 0$ geometrically?
3. What additional constraints would enforce $a + b < 0$?
4. How would the solution change if $a$ and $b$ are restricted to integers?
5. What happens if we relax the condition to $(a+b)^2 \leq a^2 + b^2$?

Tip: Always simplify inequalities by canceling like terms carefully to reveal the core relationship.