Let's examine the given statement and prove it by reasoning with equivalences.

Problem Statement

We have two positive real numbers $a$ and $b$ such that: $a + b + 2 = 2\sqrt{a} + 2\sqrt{b}$ We need to show that this equation implies: $a = b = 1$

Solution

Let's proceed by simplifying and analyzing the given equation step-by-step.

1. Rewrite the Equation: $a + b + 2 = 2\sqrt{a} + 2\sqrt{b}$ Divide both sides by 2 to simplify: $\frac{a}{2} + \frac{b}{2} + 1 = \sqrt{a} + \sqrt{b}$

2. Introduce New Variables: Let $x = \sqrt{a}$ and $y = \sqrt{b}$, where $x, y > 0$. Then $a = x^2$ and $b = y^2$, and we can rewrite the equation as: $x^2 + y^2 + 2 = 2x + 2y$ Simplifying further: $x^2 - 2x + y^2 - 2y + 2 = 0$

3. Complete the Square:

   • For the $x$-terms: $x^2 - 2x = (x - 1)^2 - 1$
   • For the $y$-terms: $y^2 - 2y = (y - 1)^2 - 1$

   Substitute these into the equation: $(x - 1)^2 - 1 + (y - 1)^2 - 1 + 2 = 0$ Simplify by combining constants: $(x - 1)^2 + (y - 1)^2 = 0$

4. Analyze the Result: Since $(x - 1)^2$ and $(y - 1)^2$ are both squares, they are non-negative. The only way their sum can be zero is if both terms are zero: $(x - 1)^2 = 0 \quad \text{and} \quad (y - 1)^2 = 0$ Therefore, $x = 1$ and $y = 1$.

5. Conclusion: Since $x = \sqrt{a} = 1$ and $y = \sqrt{b} = 1$, we have $a = 1$ and $b = 1$.

Final Answer

Thus, we have shown that $a = b = 1$ is the only solution that satisfies the equation: $a + b + 2 = 2\sqrt{a} + 2\sqrt{b} \iff a = b = 1$

Would you like further details on any step?

Here are some related questions that expand on this topic:

1. How does completing the square help in solving equations involving squares?
2. Why is it necessary to use variables like $x$ and $y$ for square roots in proofs?
3. What are alternative methods to solve similar equations without completing the square?
4. Can you apply similar reasoning to inequalities, such as $a + b + 2 > 2\sqrt{a} + 2\sqrt{b}$?
5. How can you verify solutions by substituting them back into the original equation?

Tip: When you encounter square root terms, substituting with variables for simplicity (like $x = \sqrt{a}$) can make it easier to solve the equation systematically.