Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy conditions:
$f(x) + f(y) \ge xy$ for all real $x, y$ and
for each real $x$ exists real $y$, such that $f(x) + f(y) = xy$.


1. **Setting \( x = y \) in the inequality:**
   \[
   f(x) + f(x) \ge x \cdot x
   \]
   Simplifying, we get:
   \[
   2f(x) \ge x^2 \implies f(x) \ge \frac{x^2}{2}
   \]

2. **Using the second condition:**
   For each real \( x \), there exists a real \( y \) such that:
   \[
   f(x) + f(y) = xy
   \]
   Let's denote this specific \( y \) as \( y_x \). Therefore, we have:
   \[
   f(x) + f(y_x) = xy_x
   \]

3. **Combining both conditions:**
   From the inequality \( f(x) + f(y) \ge xy \) and the equality \( f(x) + f(y_x) = xy_x \), we can infer that:
   \[
   f(x) + f(y_x) = xy_x \implies f(x) + f(y_x) \ge xy_x
   \]
   This implies that the inequality holds as an equality for the specific \( y_x \).

4. **Analyzing the equality condition:**
   Since \( f(x) \ge \frac{x^2}{2} \) and \( f(y_x) \ge \frac{y_x^2}{2} \), we have:
   \[
   f(x) + f(y_x) \ge \frac{x^2}{2} + \frac{y_x^2}{2}
   \]
   Given that \( f(x) + f(y_x) = xy_x \), we get:
   \[
   xy_x \ge \frac{x^2}{2} + \frac{y_x^2}{2}
   \]

5. **Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:**
   By AM-GM, we know:
   \[
   \frac{x^2}{2} + \frac{y_x^2}{2} \ge \sqrt{\frac{x^2}{2} \cdot \frac{y_x^2}{2}} = \frac{xy_x}{2}
   \]
   Therefore:
   \[
   xy_x \ge \frac{xy_x}{2}
   \]
   This implies:
   \[
   xy_x \ge \frac{xy_x}{2} \implies xy_x \ge 0
   \]

6. **Equality condition in AM-GM:**
   The equality in AM-GM holds if and only if \( x^2 = y_x^2 \). Therefore, \( y_x = x \) or \( y_x = -x \).

7. **Considering \( y_x = x \):**
   If \( y_x = x \), then:
   \[
   f(x) + f(x) = x \cdot x \implies 2f(x) = x^2 \implies f(x) = \frac{x^2}{2}
   \]

8. **Considering \( y_x = -x \):**
   If \( y_x = -x \), then:
   \[
   f(x) + f(-x) = x \cdot (-x) = -x^2
   \]
   Since \( f(x) \ge \frac{x^2}{2} \) and \( f(-x) \ge \frac{x^2}{2} \), we have:
   \[
   f(x) + f(-x) \ge \frac{x^2}{2} + \frac{x^2}{2} = x^2
   \]
   This contradicts \( f(x) + f(-x) = -x^2 \). Therefore, \( y_x = -x \) is not possible.

9. **Conclusion:**
   The only function that satisfies both conditions is:
   \[
   f(x) = \frac{x^2}{2}
   \]

The final answer is \( \boxed{ f(x) = \frac{x^2}{2} } \)