Determine all pairs $(c, d) \in \mathbb{R}^2$ of real constants such that there is a sequence $(a_n)_{n\geq1}$ of positive real numbers such that, for all $n \geq 1$, $$a_n \geq c \cdot a_{n+1} + d \cdot \sum_{1 \leq j < n} a_j .$$
1. **Initial Considerations:**
   - If \( c, d < 0 \), then any sequence of positive reals satisfies the given inequality.
   - If \( c \leq 0 \), then \( a_n = \theta^n \) for sufficiently large \(\theta > 1\) satisfies the inequality.
   - If \( d \leq 0 \), then \( a_n = \left(\frac{1}{2c}\right)^n \) works.

2. **Quadrant Analysis:**
   - The only quadrant potentially not completely contained in our set is \(\{(c,d) \in \mathbb{R}^2 \mid c,d > 0\}\).

3. **Case \( c \geq 1 \):**
   - If \( c \geq 1 \), then \( a_n \geq c a_{n+1} \).
   - **Subcase \( c > 1 \):**
     - If \( c > 1 \), then \(\lim_{n \to \infty} a_n = 0\).
     - Since \( a_n \geq d \sum_{j < n} a_j \), this implies \( a_n \equiv 0 \), which contradicts \( a_n > 0 \).
     - Therefore, no point \((c,d)\) with \( c > 1, d > 0 \) is in our set.
   - **Subcase \( c = 1 \):**
     - If \( c = 1 \), the sequence is non-increasing, and \(\lim a_n = L < +\infty\).
     - If \( L = 0 \), the analysis as before applies.
     - If \( L > 0 \), then all \( a_n > c_0 \).
     - The condition \( a_n > d \sum_{j < n} a_j > d c_0 (n-1) \) contradicts our hypothesis.

4. **Case \( c \in (0,1), d > 0 \):**
   - Let \( b_n = \frac{a_n}{\sum_{j < n} a_j} \).
   - We claim that \( a_i \) is a sequence of positive reals if and only if \( b_n > 0 \) for all \( n > 0 \), and \( a_1 > 0 \).
   - From the recurrence inequality, we obtain:
     \[
     b_n \geq c \cdot \frac{a_{n+1}}{\sum_{j < n+1} a_j} \cdot \frac{\sum_{j < n+1} a_j}{\sum_{j < n} a_j} + d = c \cdot b_{n+1} (1 + b_n) + d
     \]
     which simplifies to:
     \[
     \frac{b_n - d}{b_n + 1} \geq c \cdot b_{n+1}
     \]
   - Since \( b_i > 0 \), the left-hand side is bounded by an absolute constant \( C > 0 \).
   - Let \( a = \limsup b_i \). For large enough \( n \), we have \( b_n < a + \varepsilon \), and for a subsequence \( m_k \), \( b_{m_k} \to a \) as \( k \to \infty \).
   - Taking \( n = m_k - 1 \) in the inequality, we get:
     \[
     1 - \frac{d+1}{a + \varepsilon + 1} > \frac{b_{m_k - 1} - d}{b_{m_k - 1} + 1} \geq c \cdot b_{m_k} > c \cdot (a - \varepsilon)
     \]
   - Taking \( \varepsilon \to 0 \), we obtain:
     \[
     1 - \frac{d+1}{a + 1} \geq c \cdot a \implies a - d \geq c \cdot a(a + 1) \implies c a^2 + (c - 1)a + d \leq 0
     \]
   - This inequality for \( a \) admits solutions if and only if:
     \[
     \Delta = (c - 1)^2 - 4cd \geq 0 \implies d \leq \frac{(c - 1)^2}{4c}
     \]
   - Hence, it is necessary that \( d \leq \frac{(c - 1)^2}{4c} \).

5. **Sufficiency:**
   - Let \( a_n = \theta^n \). The condition holds if:
     \[
     c \theta + \frac{d}{\theta - 1} \leq 1
     \]
   - Solving the inequality, there is such a \( \theta \) if:
     \[
     d + 1 \leq \frac{(c + 1)^2}{4c} \implies d \leq \frac{(c - 1)^2}{4c}
     \]

The final answer is \( \boxed{ (c,d) } \) in all quadrants except the positive, or \((c,d)\) in the positive quadrant with \( c \in (0,1) \) and \( d \leq \frac{(c-1)^2}{4c} \).