Problem 1671

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