Real numbers $x, y$ satisfy the inequality $x^2 + y^2 \le 2$. Orove that $xy + 3 \ge 2x + 2y$
1. Given the inequality \(x^2 + y^2 \le 2\), we need to prove that \(xy + 3 \ge 2x + 2y\).

2. Let \(xy = a\). We start by analyzing the given condition:
   \[
   x^2 + y^2 \le 2
   \]
   Since \(x^2 + y^2 \ge 0\), we have:
   \[
   0 \le x^2 + y^2 \le 2
   \]

3. Consider the expression \((x+y)^2\):
   \[
   (x+y)^2 = x^2 + y^2 + 2xy
   \]
   Using the given condition \(x^2 + y^2 \le 2\), we get:
   \[
   (x+y)^2 = x^2 + y^2 + 2a \le 2 + 2a
   \]
   Since \((x+y)^2 \ge 0\), we have:
   \[
   0 \le 2 + 2a \implies a \ge -1
   \]

4. Now, we need to show that \(xy + 3 \ge 2x + 2y\). Using \(xy = a\), we rewrite the inequality as:
   \[
   a + 3 \ge 2x + 2y
   \]

5. To proceed, we use the fact that \(a \ge -1\):
   \[
   a + 1 \ge 0 \implies a + 1 \ge 0
   \]

6. We can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality to further analyze the expression. By AM-GM inequality:
   \[
   \frac{x^2 + y^2}{2} \ge \sqrt{x^2 y^2} \implies \frac{x^2 + y^2}{2} \ge |xy| \implies x^2 + y^2 \ge 2|xy|
   \]
   Given \(x^2 + y^2 \le 2\), we have:
   \[
   2 \ge 2|xy| \implies 1 \ge |xy| \implies |a| \le 1
   \]

7. Since \(a \ge -1\) and \(|a| \le 1\), we have:
   \[
   -1 \le a \le 1
   \]

8. Now, consider the expression \(2\sqrt{2(a+1)}\):
   \[
   2\sqrt{2(a+1)} \ge 2(x+y)
   \]
   Since \(a \ge -1\), we have:
   \[
   2(a+1) \ge 0 \implies \sqrt{2(a+1)} \ge 0
   \]

9. Therefore, we can conclude:
   \[
   xy + 3 = a + 3 \ge 2(x+y)
   \]

The final answer is \( \boxed{ xy + 3 \ge 2x + 2y } \)