1)
proove for positive $a, b, c, d$ 
$ \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}  \ge 2$
To prove the inequality for positive \(a, b, c, d\):
\[
\frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b} \ge 2
\]

1. **Introduce the variables and expressions:**
   Let:
   \[
   P = \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}
   \]
   \[
   Q = \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} + \frac{a}{a+b}
   \]
   \[
   R = \frac{c}{b+c} + \frac{d}{c+d} + \frac{a}{d+a} + \frac{b}{a+b}
   \]

2. **Sum of \(Q\) and \(R\):**
   Notice that:
   \[
   Q + R = \left( \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} + \frac{a}{a+b} \right) + \left( \frac{c}{b+c} + \frac{d}{c+d} + \frac{a}{d+a} + \frac{b}{a+b} \right)
   \]
   Each fraction in \(Q\) and \(R\) sums to 1:
   \[
   \frac{b}{b+c} + \frac{c}{b+c} = 1, \quad \frac{c}{c+d} + \frac{d}{c+d} = 1, \quad \frac{d}{d+a} + \frac{a}{d+a} = 1, \quad \frac{a}{a+b} + \frac{b}{a+b} = 1
   \]
   Therefore:
   \[
   Q + R = 4
   \]

3. **Sum of \(P\) and \(Q\):**
   Consider:
   \[
   P + Q = \left( \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b} \right) + \left( \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} + \frac{a}{a+b} \right)
   \]
   Grouping terms:
   \[
   P + Q = \frac{a+b}{b+c} + \frac{b+c}{c+d} + \frac{c+d}{d+a} + \frac{d+a}{a+b}
   \]
   By the Nesbitt's inequality for four variables, we have:
   \[
   \frac{a+b}{b+c} + \frac{b+c}{c+d} + \frac{c+d}{d+a} + \frac{d+a}{a+b} \ge 4
   \]
   Thus:
   \[
   P + Q \ge 4
   \]

4. **Sum of \(P\) and \(R\):**
   Consider:
   \[
   P + R = \left( \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b} \right) + \left( \frac{c}{b+c} + \frac{d}{c+d} + \frac{a}{d+a} + \frac{b}{a+b} \right)
   \]
   Grouping terms:
   \[
   P + R = \frac{a+c}{b+c} + \frac{b+d}{c+d} + \frac{c+a}{d+a} + \frac{d+b}{a+b}
   \]
   By the Nesbitt's inequality for four variables, we have:
   \[
   \frac{a+c}{b+c} + \frac{b+d}{c+d} + \frac{c+a}{d+a} + \frac{d+b}{a+b} \ge 4
   \]
   Thus:
   \[
   P + R \ge 4
   \]

5. **Combining the inequalities:**
   We have:
   \[
   P + Q \ge 4 \quad \text{and} \quad P + R \ge 4
   \]
   Adding these two inequalities:
   \[
   (P + Q) + (P + R) \ge 4 + 4
   \]
   Simplifying:
   \[
   2P + Q + R \ge 8
   \]
   Since \(Q + R = 4\), we substitute:
   \[
   2P + 4 \ge 8
   \]
   Solving for \(P\):
   \[
   2P \ge 4 \implies P \ge 2
   \]

Thus, we have proven that:
\[
\frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b} \ge 2
\]
\(\blacksquare\)

The final answer is \( \boxed{ \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b} \ge 2 } \)