We are given the implication to prove:

Statement: If $x + y \geq 2$, then $x \geq 1$ or $y \geq 1$, where $x, y \in \mathbb{R}$.

This is a disjunction proof (an "or" statement), so we need to show that whenever $x + y \geq 2$, at least one of the conditions $x \geq 1$ or $y \geq 1$ must hold.

Proof by Contrapositive:

We will prove this by proving the contrapositive of the statement. The contrapositive of the statement is logically equivalent to the original statement. The contrapositive of "If $x + y \geq 2$, then $x \geq 1$ or $y \geq 1$" is:

Contrapositive Statement: If $x < 1$ and $y < 1$, then $x + y < 2$.

If we can prove this contrapositive, the original statement will also be proven.

Step 1: Assume $x < 1$ and $y < 1$.

This means both $x$ and $y$ are less than 1.

Step 2: Add these inequalities together.

$x + y < 1 + 1 = 2$

Therefore, $x + y < 2$.

Conclusion:

We have shown that if $x < 1$ and $y < 1$, then $x + y < 2$. This proves the contrapositive of the original statement. Hence, the original statement "If $x + y \geq 2$, then $x \geq 1$ or $y \geq 1$" is true.

Summary:

We have proved that if $x + y \geq 2$, then at least one of $x \geq 1$ or $y \geq 1$ must hold, by showing that if neither condition holds (i.e., if both $x < 1$ and $y < 1$), then $x + y < 2$, which contradicts the hypothesis $x + y \geq 2$.

Would you like further details on this proof?

Here are some related questions for practice:

1. Prove that if $x + y = 2$, then $x = 1$ or $y = 1$.
2. Show that if $x + y > 2$, then $x > 1$ or $y > 1$.
3. What happens if $x + y \geq 0$? Can we still conclude something similar?
4. Prove that if $x \geq 0$ and $y \geq 0$, then $x + y \geq 2$ implies $x \geq 1$ or $y \geq 1$.
5. Prove that if $x + y < 2$, then $x < 1$ and $y < 1$.

Tip: When proving statements involving "or" (disjunctions), considering the contrapositive can often simplify the proof process.