For each $ k$ which is odd, we have $ na_{1}^k \geq \sum_{i = 1}^{n}{a_{i}^k} \geq 0$, therefore ${ a_{1} \geq 0}$ + If $ a_{n} \geq 0 \rightarrow p = a_{1}$ + If $ a_{n} < 0 \rightarrow - a_{n}^k \leq \sum_{i = 1}^{n - 1}{a_{i}^k} \leq (n - 1)a_{1}^k \rightarrow - a_{n} \leq (n - 1)^{\frac {1}{k}}a_{1}$ Because $ \lim_{k \to \infty}{(n - 1)^{\frac {1}{k}} = 1}$, therefore $ - a_{n} \leq a_{1} \rightarrow a_{1} = p$ With $ k = 1$, we get $ a_{1} \geq - \sum_{i = 2}^{n}{a_{i}}$ Applying AM-GM: $ (x - a_2)\cdot(x - a_3)\cdots (x - a_n) \leq (\frac {(n - 1)x - \sum_{i = 2}^{n}{a_i}}{n - 1})^{n - 1} \leq (x + \frac {a_1}{n - 1})^{n - 1}$ But $ (x + \frac {a_1}{n - 1})^{n - 1} = x^{n - 1} + \sum_{i = 2}^{n - 2}{\binom{n}{i}x^{n - i}(\frac {a_1}{n - 1})^i} + a_1^{n - 1} \leq x^{n - 1} + \sum_{i = 2}^{n - 2}{x^{n - i}a_1^i} + a_1^{n - 1} = \frac {x^n - a_1^n}{x - a_1}$ (because $ \binom{n}{i}\cdot\frac {1}{(n - 1)^i} < 1$) So $ (x - a_1) \cdot (x - a_2) \cdots (x - a_n) \leq x^n - a^n_1$

$(x - a_1) \cdot (x - a_2) \cdots (x - a_n) = x^n - a^n_1\iff (n=1$ or $n=2\wedge a_2=-a_1)$

Sorry to revive the post. I used induction instead.

WolfusA wrote: $(x - a_1) \cdot (x - a_2) \cdots (x - a_n) = x^n - a^n_1\iff (n=1$ or $n=2\wedge a_2=-a_1)$ sorry but what do you mean?

Condition for equality in the inequality is $$n=1\vee (n=2\wedge a_1=-a_2)$$

joepro wrote: But $ (x + \frac {a_1}{n - 1})^{n - 1} = x^{n - 1} + \sum_{i = 2}^{n - 2}{\binom{n}{i}x^{n - i}(\frac {a_1}{n - 1})^i} + a_1^{n - 1} \leq x^{n - 1} + \sum_{i = 2}^{n - 2}{x^{n - i}a_1^i} + a_1^{n - 1} = \frac {x^n - a_1^n}{x - a_1}$ I think you mean: $ (x + \frac {a_1}{n - 1})^{n - 1} = x^{n - 1} + \sum_{i = 2}^{n - 1}{\binom{n}{i}x^{n - i}(\frac {a_1}{n - 1})^i}\leq ...$ since the last term is $\displaystyle(\frac{a_1}{n-1})^{n-1}$. But otherwise nice solution!

For the first part, it is easy to see that $k\to \infty$ for some $k,$ if the largest number $p$ is actually negative, then the first expression will not be true. For the second part, could somebody help with me with the induction step. I managed to prove that if we assume that $n$ is true, then we just want to show that $a_{n+1}(x^n-a_1^n)+a_1^n(x-a_1)\geq 0$ is true, but I am not sure how to proceed from here.

very nice problem , Solved with twbnrftw Solution: 1st part we observe that not all of the terms can be negative so at least one of the terms is positive, now since $a_1$ is the maximum of all $a_{i}'s$ we have $a_{1} \geqslant 0$, now we also have for each odd integer $k$ , $\sum_{i=1}^{n} a_{i}^{k} \leqslant na_{1}^{k}$ , now we claim that $\max{\{|a_{1}|, |a_{2}|, \cdots, |a_{n}|\}}=a_{1}$. This is equivalent to show that $\max{\{|a_{1}|, |a_{2}|, \cdots , |a_{n}|\}} \neq |a_{n}|$ , now FTSOC let us assume that is the case (which can only happen if $a_{n}<0$) , then we have for sufficiently large odd $k$, $a_{1}^{k}+a_{2}^{k}+\cdots+a_{n-1}^{k}+a_{n}^{k} \leqslant (n-1)a_{1}^{k}+a_{n}^{k} $ , now we show that for sufficiently large choosen $k$ we have $(n-1)a_{1}^{k}+a_{n}^{k} <0$, but that is actually true because if we had $(n-1) \geqslant -\frac{a_{n}}{a_{1}}^{k}$ for large such odd $k$ we have $n-1 \rightarrow \infty$ which is not possible since $n$ is fixed , now that implies for large enough odd $k$ we have $a_{1}^{k} +a_{2}^{k}+ \cdots +a_{n}^{k} <0$ contradiction, hence $\max{\{|a_{1}|, |a_{2}|, \cdots , |a_{n}|\}}=a_{1}$. $\square$ Now we work on the second part clearly for for $x>a_{1}$ we have $x-a_{i}>0$ $\forall$ $i \in \{1,2, \cdots , n\}$ , hence by $\emph{AM} \geqslant \emph{GM}$ we have $(x-a_{2})(x-a_{3})\cdots (x-a_{n}) \leqslant \left( x -\frac{\sum_{i=2}^{i=n} a_{i}}{n-1}\right)^{n-1}$ , also for $k=1$ we have $a_{1} \geqslant -(a_2+a_3+\cdots +a_n) \implies \left( x -\frac{\sum_{i=2}^{i=n} a_{i}}{n-1}\right)^{n-1} \leqslant \left(x+\frac{a_{1}}{n-1}\right)^{n-1}$ , Now: $\left(x+\frac{a_{1}}{n-1}\right)^{n-1} =x^{n-1}+ a_{1}\frac{\binom{n-1}{1}}{n-1}+a_{2}^{2}\frac{\binom{n-1}{2}}{(n-1)^2}+\cdots +a_{1}^{n-1}< x^{n-1}+x^{n-2}a_{1}+x^{n-3}a_{2}^{2}+\cdots +a_{1}^{n-1}=\frac{x-a_{1}^{n}}{x-a_{1}} \implies (x-a_{1})(x-a_{2})(x-a_{3})\cdots (x-a_{n}) \leqslant x^{n}-a_{1}^{n}$ , equality holds at $n=1$ or $n=2$ with $a_{2}=-a_{1}$. $\square$