Find the largest interval $ M \subseteq \mathbb{R^ \plus{} }$, such that for all $ a$, $ b$, $ c$, $ d \in M$ the inequality
\[ \sqrt {ab} \plus{} \sqrt {cd} \ge \sqrt {a \plus{} b} \plus{} \sqrt {c \plus{} d}\]
holds. Does the inequality
\[ \sqrt {ab} \plus{} \sqrt {cd} \ge \sqrt {a \plus{} c} \plus{} \sqrt {b \plus{} d}\]
hold too for all $ a$, $ b$, $ c$, $ d \in M$?
($ \mathbb{R^ \plus{} }$ denotes the set of positive reals.)
To find the largest interval \( M \subseteq \mathbb{R^+} \) such that for all \( a, b, c, d \in M \), the inequality
\[ \sqrt{ab} + \sqrt{cd} \ge \sqrt{a + b} + \sqrt{c + d} \]
holds, we start by testing specific values and then generalize.

1. **Testing specific values:**
   Let \( a = b = c = d \). Then the inequality becomes:
   \[ 2\sqrt{a^2} \ge 2\sqrt{2a} \]
   Simplifying, we get:
   \[ 2a \ge 2\sqrt{2a} \]
   Dividing both sides by 2:
   \[ a \ge \sqrt{2a} \]
   Squaring both sides:
   \[ a^2 \ge 2a \]
   Rearranging:
   \[ a^2 - 2a \ge 0 \]
   Factoring:
   \[ a(a - 2) \ge 0 \]
   Since \( a \in \mathbb{R^+} \), we have:
   \[ a \ge 2 \]
   Therefore, \( M \subseteq [2, \infty) \).

2. **Generalizing the inequality:**
   We need to show that \( M = [2, \infty) \) works for all \( a, b, c, d \in M \). Squaring the given inequality:
   \[ (\sqrt{ab} + \sqrt{cd})^2 \ge (\sqrt{a + b} + \sqrt{c + d})^2 \]
   Expanding both sides:
   \[ ab + cd + 2\sqrt{abcd} \ge a + b + c + d + 2\sqrt{(a + b)(c + d)} \]
   We need to show:
   \[ ab + cd + 2\sqrt{abcd} \ge a + b + c + d + 2\sqrt{(a + b)(c + d)} \]
   Since \( a, b, c, d \ge 2 \), we can use the fact that:
   \[ ab \ge a + b \quad \text{and} \quad cd \ge c + d \]
   This follows from the inequality \( ab - a - b = (a - 1)(b - 1) - 1 \ge 0 \) for \( a, b \ge 2 \).

3. **Verifying the second inequality:**
   We need to check if the inequality:
   \[ \sqrt{ab} + \sqrt{cd} \ge \sqrt{a + c} + \sqrt{b + d} \]
   holds for all \( a, b, c, d \in M \).

   Using the same approach, let \( a = b = c = d \). Then:
   \[ 2\sqrt{a^2} \ge 2\sqrt{2a} \]
   Simplifying, we get:
   \[ 2a \ge 2\sqrt{2a} \]
   Dividing both sides by 2:
   \[ a \ge \sqrt{2a} \]
   Squaring both sides:
   \[ a^2 \ge 2a \]
   Rearranging:
   \[ a^2 - 2a \ge 0 \]
   Factoring:
   \[ a(a - 2) \ge 0 \]
   Since \( a \in \mathbb{R^+} \), we have:
   \[ a \ge 2 \]
   Therefore, \( M \subseteq [2, \infty) \).

   Thus, the second inequality also holds for \( M = [2, \infty) \).

The final answer is \( \boxed{ M = [2, \infty) } \).