Let $n\ge 2$ be a positive integer. There are $n$ real coefficient polynomials $P_1(x),P_2(x),\cdots ,P_n(x)$ which is not all the same, and their leading coefficients are positive. Prove that
$$\deg(P_1^n+P_2^n+\cdots +P_n^n-nP_1P_2\cdots P_n)\ge (n-2)\max_{1\le i\le n}(\deg P_i)$$
and find when the equality holds.
1. **Assume the degrees of the polynomials are equal:**
   Let \( \deg(P_i) = m \) for all \( i = 1, 2, \ldots, n \). Denote the leading coefficient of \( P_i \) by \( a_i \), where \( a_i > 0 \).

2. **Consider the leading term of the polynomial:**
   The leading term of \( P_i^n \) is \( a_i^n x^{mn} \). Therefore, the leading term of \( P_1^n + P_2^n + \cdots + P_n^n \) is \( \sum_{i=1}^n a_i^n x^{mn} \).

3. **Consider the leading term of the product:**
   The leading term of \( P_1 P_2 \cdots P_n \) is \( (a_1 a_2 \cdots a_n) x^{mn} \). Therefore, the leading term of \( n P_1 P_2 \cdots P_n \) is \( n (a_1 a_2 \cdots a_n) x^{mn} \).

4. **Compare the leading coefficients:**
   The leading coefficient of \( P_1^n + P_2^n + \cdots + P_n^n - n P_1 P_2 \cdots P_n \) is \( \sum_{i=1}^n a_i^n - n (a_1 a_2 \cdots a_n) \).

5. **Use the AM-GM inequality:**
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   \frac{a_1^n + a_2^n + \cdots + a_n^n}{n} \geq (a_1 a_2 \cdots a_n)
   \]
   Multiplying both sides by \( n \), we get:
   \[
   a_1^n + a_2^n + \cdots + a_n^n \geq n (a_1 a_2 \cdots a_n)
   \]
   Therefore, \( \sum_{i=1}^n a_i^n - n (a_1 a_2 \cdots a_n) \geq 0 \).

6. **Equality condition:**
   The equality holds if and only if \( a_1 = a_2 = \cdots = a_n \).

7. **Consider the polynomials \( P_i(x) \):**
   Assume \( P_i(x) = a_i x^m + Q_i(x) \) where \( \deg(Q_i) < m \). If \( a_i = 1 \) for all \( i \), then \( P_i(x) = x^m + Q_i(x) \).

8. **Expand \( P_i(x)^n \) and the product:**
   \[
   P_i(x)^n = (x^m + Q_i(x))^n = x^{mn} + \sum_{k=1}^n \binom{n}{k} x^{m(n-k)} Q_i(x)^k
   \]
   \[
   \prod_{i=1}^n (x^m + Q_i(x)) = x^{mn} + \sum_{k=1}^n x^{m(n-k)} \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq n} Q_{i_1} Q_{i_2} \cdots Q_{i_k}
   \]

9. **Compare the coefficients:**
   The coefficient of \( x^{m(n-1)} \) in both expansions is:
   \[
   n \sum_{i=1}^n x^{m(n-1)} Q_i(x) - n \sum_{i=1}^n x^{m(n-1)} Q_i(x) = 0
   \]

10. **Consider the next highest degree term:**
    Let \( r = \max_{1 \leq i \leq n} (\deg Q_i) \) and \( q_i \) be the coefficient of \( x^r \) in \( Q_i \). The coefficient of \( x^{m(n-2) + 2r} \) in the polynomial \( P \) is:
    \[
    \binom{n}{2} \sum_{i=1}^n q_i^2 - n \sum_{1 \leq i_1 < i_2 \leq n} q_{i_1} q_{i_2}
    \]
    This can be written as:
    \[
    \frac{n}{2} \left( (n-1) \sum_{i=1}^n q_i^2 - \sum_{1 \leq i_1 < i_2 \leq n} q_{i_1} q_{i_2} \right)
    \]
    The expression in the brackets is non-negative and equals zero only when \( q_i \) are all equal.

11. **Determine the degree of \( P \):**
    Let \( r_0 \) be the greatest integer such that the coefficients of \( x^{r_0} \) in \( Q_i \) are not all equal. The coefficient of \( x^{m(n-2) + 2r_0} \) in \( P \) is positive. Therefore:
    \[
    \deg(P) \geq m(n-2) + r_0 \geq (n-2)m
    \]
    The equality holds when \( r_0 = 0 \), i.e., when \( Q_i(x) = Q(x) + c_i \) and \( c_i \in \mathbb{R} \) are not all equal.

The final answer is \( \boxed{ \deg(P_1^n + P_2^n + \cdots + P_n^n - n P_1 P_2 \cdots P_n) \geq (n-2) \max_{1 \leq i \leq n} (\deg P_i) } \)