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The exercise asks to show that for all \( x \geq 1 \) and \( y \geq 1 \), the following equation:

\[
x^2 + y^2 + xy - x - y - 1 = 0
\]

implies that \( x = y = 1 \).

### Step-by-Step Solution:

#### Step 1: Evaluate the equation at \( x = 1 \) and \( y = 1 \)

We first check if the equation holds when \( x = 1 \) and \( y = 1 \):

\[
1^2 + 1^2 + 1 \cdot 1 - 1 - 1 - 1 = 0.
\]

Simplifying the expression:

\[
1 + 1 + 1 - 1 - 1 - 1 = 0.
\]

Thus, the equation is true for \( x = 1 \) and \( y = 1 \).

#### Step 2: Analyze for \( x > 1 \) and \( y > 1 \)

Now, consider \( x > 1 \) and \( y > 1 \). We need to check if the equation holds for any values other than \( x = 1 \) and \( y = 1 \).

Rearrange the equation:

\[
x^2 + y^2 + xy = x + y + 1.
\]

For \( x > 1 \) and \( y > 1 \), each of the terms \( x^2 \), \( y^2 \), and \( xy \) will grow faster than the linear terms \( x + y + 1 \). Hence, the left-hand side (LHS) will be greater than the right-hand side (RHS), which means the equation cannot be satisfied for \( x > 1 \) and \( y > 1 \).

Thus, the only solution where the equation holds is \( x = y = 1 \).

### Conclusion:
The given equation holds true only for \( x = y = 1 \) when both \( x \geq 1 \

Show that for all x ≥ 1 and y ≥ 1, the equation x^2 + y^2 + xy - x - y - 1 = 0 implies that x = y = 1.

Explore how to prove that the equation x^2 + y^2 + xy - x - y - 1 = 0 implies x = y = 1 for values of x ≥ 1 and y ≥ 1. This problem involves key algebraic concepts such as quadratic terms and inequalities, suitable for students in Grades 10-12.

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