Let $a$, $b$ and $c$ positive integers. Three sequences are defined as follows:
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[*] $a_1=a$, $b_1=b$, $c_1=c$[/*]
[*] $a_{n+1}=\lfloor{\sqrt{a_nb_n}}\rfloor$, $\:b_{n+1}=\lfloor{\sqrt{b_nc_n}}\rfloor$, $\:c_{n+1}=\lfloor{\sqrt{c_na_n}}\rfloor$ for $n \ge 1$[/*]
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[list = a]
[*]Prove that for any $a$, $b$, $c$, there exists a positive integer $N$ such that $a_N=b_N=c_N$.[/*]
[*]Find the smallest $N$ such that $a_N=b_N=c_N$ for some choice of $a$, $b$, $c$ such that $a \ge 2$ y $b+c=2a-1$.[/*]
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### Part (a)
1. Define the sequences \(a_n\), \(b_n\), and \(c_n\) as given:
   \[
   a_1 = a, \quad b_1 = b, \quad c_1 = c
   \]
   \[
   a_{n+1} = \lfloor \sqrt{a_n b_n} \rfloor, \quad b_{n+1} = \lfloor \sqrt{b_n c_n} \rfloor, \quad c_{n+1} = \lfloor \sqrt{c_n a_n} \rfloor \quad \text{for} \quad n \geq 1
   \]

2. Consider the sum \(S_n = a_n + b_n + c_n\). We need to show that \(S_n\) is non-increasing:
   \[
   S_{n+1} = a_{n+1} + b_{n+1} + c_{n+1}
   \]
   Using the definition of the sequences:
   \[
   S_{n+1} = \lfloor \sqrt{a_n b_n} \rfloor + \lfloor \sqrt{b_n c_n} \rfloor + \lfloor \sqrt{c_n a_n} \rfloor
   \]

3. Since \(\lfloor x \rfloor \leq x\) for any real number \(x\), we have:
   \[
   \lfloor \sqrt{a_n b_n} \rfloor \leq \sqrt{a_n b_n}, \quad \lfloor \sqrt{b_n c_n} \rfloor \leq \sqrt{b_n c_n}, \quad \lfloor \sqrt{c_n a_n} \rfloor \leq \sqrt{c_n a_n}
   \]
   Therefore:
   \[
   S_{n+1} \leq \sqrt{a_n b_n} + \sqrt{b_n c_n} + \sqrt{c_n a_n}
   \]

4. By the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality):
   \[
   \sqrt{a_n b_n} \leq \frac{a_n + b_n}{2}, \quad \sqrt{b_n c_n} \leq \frac{b_n + c_n}{2}, \quad \sqrt{c_n a_n} \leq \frac{c_n + a_n}{2}
   \]
   Summing these inequalities:
   \[
   \sqrt{a_n b_n} + \sqrt{b_n c_n} + \sqrt{c_n a_n} \leq \frac{a_n + b_n}{2} + \frac{b_n + c_n}{2} + \frac{c_n + a_n}{2} = a_n + b_n + c_n = S_n
   \]

5. Thus, we have:
   \[
   S_{n+1} \leq S_n
   \]
   Since \(S_n\) is a non-increasing sequence of positive integers, it must eventually stabilize. Let \(N\) be the smallest integer such that \(S_{N+1} = S_N\).

6. If \(S_{N+1} = S_N\), then:
   \[
   \sqrt{a_N b_N} + \sqrt{b_N c_N} + \sqrt{c_N a_N} = a_N + b_N + c_N
   \]
   This equality holds if and only if \(a_N = b_N = c_N\). Therefore, there exists a positive integer \(N\) such that \(a_N = b_N = c_N\).

### Part (b)
1. We need to find the smallest \(N\) such that \(a_N = b_N = c_N\) for some choice of \(a\), \(b\), and \(c\) such that \(a \geq 2\) and \(b + c = 2a - 1\).

2. Consider the example \(a = 2\), \(b = 2\), and \(c = 1\):
   \[
   a_1 = 2, \quad b_1 = 2, \quad c_1 = 1
   \]
   \[
   a_2 = \lfloor \sqrt{2 \cdot 2} \rfloor = \lfloor \sqrt{4} \rfloor = 2
   \]
   \[
   b_2 = \lfloor \sqrt{2 \cdot 1} \rfloor = \lfloor \sqrt{2} \rfloor = 1
   \]
   \[
   c_2 = \lfloor \sqrt{1 \cdot 2} \rfloor = \lfloor \sqrt{2} \rfloor = 1
   \]

3. For \(N = 3\):
   \[
   a_3 = \lfloor \sqrt{2 \cdot 1} \rfloor = \lfloor \sqrt{2} \rfloor = 1
   \]
   \[
   b_3 = \lfloor \sqrt{1 \cdot 1} \rfloor = \lfloor \sqrt{1} \rfloor = 1
   \]
   \[
   c_3 = \lfloor \sqrt{1 \cdot 2} \rfloor = \lfloor \sqrt{2} \rfloor = 1
   \]

4. Thus, \(a_3 = b_3 = c_3 = 1\). Therefore, the smallest \(N\) such that \(a_N = b_N = c_N\) for the given conditions is \(N = 3\).

The final answer is \( \boxed{ N = 3 } \)