It was delirium. It seems the following is also true. $$\deg(P_1^{n-1}P_2+P_2^{n-1}P_3+\cdots +P_n^{n-1}P_1-nP_1P_2\cdots P_n)\ge (n-2)\max_{1\le i\le n}(\deg P_i)$$

arqady wrote: Let $\Phi$ be cyclic homogeneous polynomial with $n$ variables, $\deg{\Phi}=n$, $\Phi(1,1,...,1)=0$ and $P_i$ as above. Prove that: $$\deg\Phi\left(P_1,P_2,...,P_n\right)\geq(n-2)\max_{1\le i\le n}(\deg P_i).$$ I doubt it can be generalized like that, only with these assumptions. F_Xavier1203 wrote: 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. It's only about comparing a few of the coefficients at the greatest powers. Denote the polynomial at the LHS as $P$. In case $P_i, i=1,2,\dots,n$ are not of the same degree, the degree of $P$ is $n\max(\deg P_i)$ and we are done, so assume $\deg P_i=m\ge 0, i=1,2,\dots,n$. Denote by $a_i>0$ the senior coefficient of $P_i$. The coefficient at $x^{mn}$ equals $$\sum_{i=1}^n a_i^n - n\prod_{i=1}^n a_i\ge 0$$(by a well known inequality). Moreover, the equality is attained only when all $a_i,i=1,\dots,n$ are equal. Further, we assume $a_i=1,i=1,\dots,n$. Let $P_i(x)=x^m + Q_i(x), \deg Q_i< m.$ We have $$P_i(x)^n=\sum_{k=0}^n \binom{n}{k}x^{m (n-k)} Q_i(x)^k \qquad(1)$$and $$\prod_{i=1}^n\left(x^m+Q_i(x)\right)=x^{mn}+\sum_{k=1}^n x^{m(n-k)}\sum_{1\le i_1<i_2<\dots<i_k\le n}Q_{i_1}Q_{i_2}\cdots Q_{i_k} \qquad(2)$$Now, putting $k=1$ in (1) and (2) yields $$n\sum_{i=1}^n x^{m(n-1)}Q_i(x)-n\sum_{i=1}^nx^{m(n-1)}Q_i(x)=0.$$Let $r:=\max_{1\le i\le n} (\deg Q_i) $ and $q_i$ be the coefficient at $x^r$ of $Q_i,i=1,\dots,n$. Let's put $k=2$ in (1) and (2). The coefficient at $x^{m(n-2)+2r}$ of the polynomial $P$ equals $$\binom{n}{2}\sum_{i=1}^n q_i^2-n\sum_{i_1,i_2\in I, i_1<i_2 }q_{i_1}q_{i_2}$$which can be written as $$\frac{n}{2}\left((n-1)\sum_{i=1}^n q_i^2 - \sum_{i_1,i_2\in [1..n]}q_{i_1}q_{i_2} \right)$$Note that the sum in the big brackets is not negative, and equals $0$ only when $q_i, i=1,\dots,n$ are equal. Let $r_0, 0\le r_0\le r$ be the greatest integer such that the coefficients at $x^{r_0}$ of $Q_i, i=1,\dots,n$ are not all equal. In the same way, we obtain that the coefficient at $x^{m(n-2)+2r_0}$ of $P$ is positive. Therefore $$\deg P\ge m(n-2)+r_0\ge (n-2)m$$The equality is attained when $r_0=0$, that is, when $Q_i(x)=Q(x)+c_i$ and $c_i\in\mathbb{R},i=1,2,\dots,n$ are not all equal.

dgrozev wrote: arqady wrote: Let $\Phi$ be cyclic homogeneous polynomial with $n$ variables, $\deg{\Phi}=n$, $\Phi(1,1,...,1)=0$ and $P_i$ as above. Prove that: $$\deg\Phi\left(P_1,P_2,...,P_n\right)\geq(n-2)\max_{1\le i\le n}(\deg P_i).$$ I doubt it can be generalized like that, only with these assumptions. You are welcome to show us your counterexample.

arqady wrote: dgrozev wrote: arqady wrote: Let $\Phi$ be cyclic homogeneous polynomial with $n$ variables, $\deg{\Phi}=n$, $\Phi(1,1,...,1)=0$ and $P_i$ as above. Prove that: $$\deg\Phi\left(P_1,P_2,...,P_n\right)\geq(n-2)\max_{1\le i\le n}(\deg P_i).$$ I doubt it can be generalized like that, only with these assumptions. You are welcome to show us your counterexample. Take $n=3$ and $\Phi(p_1,p_2,p_3):=(p_1-p_2)(p_2-p_3)(p_3-p_1)$ and put $p_1=p_2=x, p_3=2x.$

dgrozev wrote: Take $n=3$ and $\Phi(p_1,p_2,p_3):=(p_1-p_2)(p_2-p_3)(p_3-p_1)$ and put $p_1=p_2=x, p_3=2x.$ Yes, you are right. Thank you!

2022 Korea Winter Program Practice Test wrote: 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. Suppose otherwise, that there exists polynomials $P_1, \dots, P_n \in \mathbb{R}[x]$, which is not all the same, with positive leading coefficient such that \[ \deg(P_1^n + \dots + P_n^n - nP_1 \cdots P_n) < (n - 2) \max_{1 \le i \le n}(\deg P_i) \]WLOG $\text{deg}(P_1) \ge \dots \ge \text{deg}(P_n)$. If $\text{deg}(P_1) > \text{deg}(P_n)$, then as all $P_i$ has positive leading coefficients, we have \[ \deg(P_1^n + \dots + P_n^n) \ge n\deg(P_1) > \deg(P_1\cdots P_n) \implies \deg(P_1^n + \dots + P_n^n - nP_1 \cdots P_n) = \deg(P_1^n + \dots + P_n^n) \ge n \deg(P_1) \]which is a contradiction. Therefore, we must have $\deg(P_1) = \dots = \deg(P_n) = m \ge 0$. Write $P_i = a_i x^m + Q_i$, where $\text{deg}(Q_i) < m$. The condition above implies that $[x^{\ell}] (P_1^n + \dots + P_n^n - nP_1 \cdots P_n) = 0$ for all $\ell \ge m(n - 2)$. \begin{align*} 0 &= [x^{mn}](P_1^n + \dots + P_n^n - nP_1 \cdots P_n) \\ &= [x^{mn}] \left( \sum_{i = 1}^n (a_i x^m + Q_i)^n - n \prod_{i = 1}^n (a_i x^m + Q_i) \right) \\ &= \sum_{i = 1}^n a_i^n - n \prod_{i = 1}^n a_i \stackrel{\text{AM-GM}}{\ge} 0 \end{align*}and thus, we must have $a_1 = a_2 = \dots = a_n = a > 0$, and $m \ge 1$ since $P_1, \dots, P_n$ are not all the same. Now, let $0 \le r \le \max_{1 \le i \le n} \deg Q_i$ be the largest positive integer such that the coefficients of $x^r$ in $P_i(x)$ is not all the same, which must exist by the hypothesis of the problem. Write $P_i = x^rR + (r_ix^r + S_i)$ for each $i$, where $R \in \mathbb{R}[x]$, $r_i \in \mathbb{R}$ and $\deg S_i < r$. Thus, $r_i$ are not all the same by our choice of $r$. First of all, note that for any $\ell \ge 3$, \begin{align*} \deg ( (x^rR)^{n - \ell} (r_ix^r + S_i)^\ell ) &= (n - \ell) \deg (x^rR) + \ell \deg (r_i x^r + S_i) \le (n - \ell)m + \ell r < m(n - 2) + 2r \\ \deg \left( \sum_{S \subseteq [n], |S| = \ell} (x^r R)^{n - \ell} \prod_{i \in S} (r_i x^r + S_i) \right) &\le \max_{1 \le i \le n} (n - \ell) \deg (x^rR) + \ell \deg(r_i x^r + S_i) \le (n - \ell)m + \ell r < m(n - 2) + 2r \end{align*}Therefore, we conclude that \begin{align*} 0 &= [x^{m(n - 2) + 2r}] (P_1^n + \dots + P_n^n - nP_1 \cdots P_n) \\ &= [x^{m(n - 2) + 2r}] \left( \sum_{i = 1}^n (x^rR + (r_ix^r + S_i))^n - n \prod_{i = 1}^n (x^rR + (r_ix^r + S_i)) \right) \\ &= [x^{m(n - 2) + 2r}] \left( \binom{n}{2} (x^rR)^{n - 2} (r_i x^r + S_i)^2 - n (x^r R)^{n - 2} \sum_{1 \le i < j \le n} (r_ix^r + S_i)(r_j x^r + S_j) \right) \\ &= \binom{n}{2} a^{n - 2} r_i^{2} - n a^{n - 2} \sum_{1 \le i < j \le n} r_i r_j \\ &= \frac{n}{2} \cdot a^{n - 2} \cdot \left( \sum_{1 \le i < j \le n} (r_i - r_j)^2 \right) \ge 0 \end{align*}where last equality hold if and only if all $r_i$s are equal, which is not the case. Therefore, this equality can't hold, and therefore our original assumption is false. Therefore, we must have \[ \deg(P_1^n + \dots + P_n^n - nP_1 \cdots P_n) \ge (n - 2) \max_{1 \le i \le n}(\deg P_i) \]Equality holds when $r = 0$, i.e. we have $P_i(x) = R(x) + r_i$ for some $r_1, \dots, r_n$ not all equal.