Suppose a function $f:\mathbb{R}\to\mathbb{R}$ satisfies $|f(x+y)|\geqslant|f(x)+f(y)|$ for all real numbers $x$ and $y$. Prove that equality always holds. Is the conclusion valid if the sign of the inequality is reversed?
1. **Assertion and Initial Conditions:**
   Let \( P(x, y) \) denote the assertion \( |f(x+y)| \geq |f(x) + f(y)| \). 
   
   - For \( P(0, 0) \):
     \[
     |f(0+0)| \geq |f(0) + f(0)| \implies |f(0)| \geq |2f(0)|
     \]
     This implies \( |f(0)| \geq 2|f(0)| \). The only solution to this inequality is \( f(0) = 0 \).

2. **Odd Function Property:**
   - For \( P(x, -x) \):
     \[
     |f(x + (-x))| \geq |f(x) + f(-x)| \implies |f(0)| \geq |f(x) + f(-x)|
     \]
     Since \( f(0) = 0 \), we have:
     \[
     0 \geq |f(x) + f(-x)|
     \]
     This implies \( f(x) + f(-x) = 0 \), hence \( f(-x) = -f(x) \). Therefore, \( f \) is an odd function.

3. **Analyzing \( P(x+y, -x) \) and \( P(x+y, -y) \):**
   - For \( P(x+y, -x) \):
     \[
     |f((x+y) + (-x))| \geq |f(x+y) + f(-x)| \implies |f(y)| \geq |f(x+y) - f(x)|
     \]
   - For \( P(x+y, -y) \):
     \[
     |f((x+y) + (-y))| \geq |f(x+y) + f(-y)| \implies |f(x)| \geq |f(x+y) - f(y)|
     \]

4. **Summing the Inequalities:**
   - Squaring both sides of the inequalities:
     \[
     f(y)^2 \geq (f(x+y) - f(x))^2 \implies f(y)^2 \geq f(x+y)^2 - 2f(x+y)f(x) + f(x)^2
     \]
     \[
     f(x)^2 \geq (f(x+y) - f(y))^2 \implies f(x)^2 \geq f(x+y)^2 - 2f(x+y)f(y) + f(y)^2
     \]
   - Adding these inequalities:
     \[
     f(x)^2 + f(y)^2 \geq f(x+y)^2 - 2f(x+y)f(x) + f(x)^2 + f(x+y)^2 - 2f(x+y)f(y) + f(y)^2
     \]
     \[
     2(f(x)^2 + f(y)^2) \geq 2f(x+y)^2 - 2f(x+y)(f(x) + f(y))
     \]
     Dividing by 2:
     \[
     f(x)^2 + f(y)^2 \geq f(x+y)^2 - f(x+y)(f(x) + f(y))
     \]
     Rearranging:
     \[
     f(x+y)(f(x) + f(y)) \geq f(x+y)^2
     \]
     This implies:
     \[
     f(x+y) \cdot (f(x) + f(y)) \geq f(x+y)^2
     \]
     Since \( f(x+y) \neq 0 \), we can divide both sides by \( f(x+y) \):
     \[
     f(x) + f(y) \geq f(x+y)
     \]

5. **Conclusion:**
   - If \( f(x+y) \geq 0 \):
     \[
     f(x) + f(y) \geq f(x+y) \geq 0
     \]
     Thus, \( |f(x) + f(y)| = f(x) + f(y) \geq f(x+y) = |f(x+y)| \).
   - If \( f(x+y) \leq 0 \):
     \[
     f(x) + f(y) \leq f(x+y) \leq 0
     \]
     Thus, \( |f(x) + f(y)| = -(f(x) + f(y)) \leq -f(x+y) = |f(x+y)| \).

Therefore, the equality \( |f(x+y)| = |f(x) + f(y)| \) always holds.

The final answer is \( \boxed{ |f(x+y)| = |f(x) + f(y)| } \).