Find all functions $f : R \to  R$ satisfying the conditions:
1. $f (x + 1) \ge  f (x) + 1$ for all $x \in R$  
2. $f (x y) \ge f (x)f (y)$ for all $x, y \in R$ 
We need to find all functions \( f : \mathbb{R} \to \mathbb{R} \) that satisfy the following conditions:
1. \( f(x + 1) \ge f(x) + 1 \) for all \( x \in \mathbb{R} \)
2. \( f(xy) \ge f(x)f(y) \) for all \( x, y \in \mathbb{R} \)

We claim that the only function that satisfies these conditions is \( \boxed{f(x) = x} \).

1. **Checking \( f(x) = x \) satisfies the conditions:**
   - For condition 1: \( f(x + 1) = x + 1 \) and \( f(x) + 1 = x + 1 \). Clearly, \( f(x + 1) = f(x) + 1 \).
   - For condition 2: \( f(xy) = xy \) and \( f(x)f(y) = xy \). Clearly, \( f(xy) = f(x)f(y) \).

2. **Proving \( f(x) = x \) is the only solution:**
   - From condition 1, \( f \) is unbounded. This means \( f(x) \) can take arbitrarily large values.
   - Let \( P(x, y) \) denote the assertion \( f(xy) \ge f(x)f(y) \).

3. **Analyzing \( P(x, x) \):**
   - \( P(x, x) \): \( f(x^2) \ge f(x)^2 \). This implies \( f(x) \ge 0 \) for all \( x \ge 0 \).

4. **Analyzing \( P(0, x) \):**
   - \( P(0, x) \): \( f(0) \ge f(0)f(x) \). This implies \( f(0)(1 - f(x)) \ge 0 \) for all \( x \).
   - Since \( f \) is unbounded, there exists \( x \) such that \( f(x) > 1 \). Together with \( f(0) \ge 0 \), we have \( f(0) = 0 \).

5. **Analyzing \( P(1, 1) \):**
   - \( P(1, 1) \): \( f(1) \ge f(1)^2 \). This implies \( f(1)(f(1) - 1) \le 0 \).
   - Since \( f(1) \ge f(0) + 1 = 1 \), we have \( f(1) = 1 \).

6. **Analyzing \( P(-1, -1) \):**
   - \( P(-1, -1) \): \( f(1) \ge f(-1)^2 \). This implies \( f(-1)^2 \le 1 \), so \( -1 \le f(-1) \le 1 \).
   - Since \( f(-1) \le f(0) - 1 = -1 \), we have \( f(-1) = -1 \).

7. **Analyzing \( P(x, -1) \):**
   - \( P(x, -1) \): \( f(-x) \ge -f(x) \) for all \( x \in \mathbb{R} \).

8. **Using condition 1:**
   - By condition 1, \( f(x + n) \ge f(x) + n \) for all \( x \in \mathbb{R} \) and \( n \in \mathbb{N} \).

9. **Using condition 2:**
   - By condition 2, \( f(x^n) \ge f(x)^n \) for all \( x > 0 \) and \( n \in \mathbb{N} \).

10. **Bounding \( f(x) \) for \( x \ge 0 \):**
    - Note that \( f(x) = f(\lfloor x \rfloor + \{ x \}) \ge f(\{ x \}) + \lfloor x \rfloor \ge \lfloor x \rfloor > x - 1 \) for all \( x \ge 0 \).

11. **Bounding \( f(x) \) for \( x \in (0, 1) \):**
    - For all \( x > 1 \), \( (x - 1)f\left(\frac{1}{x}\right) < f(x)f\left(\frac{1}{x}\right) \le 1 \). This implies \( f\left(\frac{1}{x}\right) < \frac{1}{x - 1} \).
    - Plugging \( x \to \frac{1}{x} \), we get \( f(x) < \frac{x}{1 - x} \) for all \( x \in (0, 1) \).

12. **Bounding \( f(x) \) for \( x \in (0, 1) \):**
    - Then \( f(x)^n \le f(x^n) < \frac{x^n}{1 - x^n} \). This implies \( f(x) < \frac{x}{\sqrt[n]{1 - x^n}} \) for all \( x \in (0, 1) \).
    - As \( n \) goes to infinity, \( \sqrt[n]{1 - x^n} \) approaches 1. Thus \( f(x) \le x \) for all \( x \in (0, 1) \).

13. **Bounding \( f(x) \) for \( x \in (-1, 0) \):**
    - \( f(-x) \ge -f(x) \) and \( -f(x) \ge -x \) for all \( x \in (0, 1) \) is equivalent to \( f(x) \ge x \) for all \( x \in (-1, 0) \).

14. **Analyzing \( f(x) \) for \( x \in [-1, 1] \):**
    - For all \( x \in (-1, 0) \), \( x = (x + 1) - 1 \ge f(x + 1) - 1 \ge f(x) \ge x \).
    - Since all equalities have to hold, \( f(x) = x \) and \( f(x + 1) = x + 1 \) which means \( f(x) = x \) for all \( x \in [-1, 1] \).

15. **Generalizing for all \( x \neq 0 \):**
    - For all \( x \neq 0 \), there exists \( c \in (0, 1) \) such that \( |cx| < 1 \).
    - \( P(x, c) \): \( cx \ge cf(x) \). This implies \( f(x) \le x \).
    - \( P(x, -c) \): \( -cx \ge -cf(x) \). This implies \( f(x) \ge x \).
    - Thus, \( f(x) = x \) for all \( x \neq 0 \), and with \( f(0) = 0 \), we conclude that \( f(x) \equiv x \) is the only solution.

\(\blacksquare\)