Polynomial inequality

I found the following problem on a website and would be curious to find a solution.

Let $a_1\ge a_2\ge\cdots\ge a_n$ be real numbers such that for all integer $k>0$: $$a_1^k+a_2^k+\cdots+a_n^k\ge 0$$ Let $p=\max\{|a_1|,\ldots,|a_n|\}$. Show that $p=a_1$, and that $$(x-a_1)(x-a_2)\cdots(x-a_n)\le x^n-a_1^n$$ for all $x>a_1$.

The first part is rather simple since we can normalize by $p$ and consider the limit $k\to\infty$. However, I can't seem to get the second part right...

POSSIBLE SOLUTION FOR $n\ge 6$: Without loss of generalization, assume that $a_1=1$. Since $x>|a_i|$ for all $i$, the following expansions converge: $$\sum_i\ln\left(1-\frac{a_i}{x}\right)=-\sum_{i}\sum_{k=1}^\infty\frac{a_i^k}{kx^k}.$$ Since the sum over $i$ for a given $k$ is non-negative, and we have $$\sum_i\ln\left(1-\frac{a_i}{x}\right)\le-\sum_{k=1}^\infty\sum_i\left(a_i^{2k}+a_i^{2k+1}\right)\frac{1}{(2k+1)x^{2k+1}}.$$ Notice that $a_i^{2k}+a_i^{2k+1}\ge0$ since $|a_i|\le1$. Then $$\sum_i\left(a_i^{2k}+a_i^{2k+1}\right)\ge2.$$ We have $$\sum_i\ln\left(1-\frac{a_i}{x}\right)\le-\sum_{k=1}^\infty\frac 2{(2k+1)x^{2k+1}}\le-\sum_{k=1}^\infty\frac{2}{(2k+2)x^{2k+2}}.$$ The last inequality then reduces to $$\sum_i\ln\left(1-\frac{a_i}{x}\right)\le-\sum_{k=2}^\infty\frac{1}{kx^{2k}}=-\sum_{k=1}^\infty\frac{1}{kx^{2k}}+\frac{1}{x^2}.$$ Therefore, $$\prod\left(x-a_i\right)\le x^n\left(1-\frac{1}{x^2}\right)\exp(1/x^2).$$ We can actually show that this is a sharper bound that the one required in the question, for $n\ge 6$. Consider the functions $f(y)=(1-y)\exp(y)$ and $g(y)=(1-y^3)$ on the interval $[0,1]$. We have $f(0)-g(0)=f(1)-g(1)=0$. Furthermore, $f'(y)-g'(y)=y(-\exp(y)+3y)$, $f'(0)-g'(0)<0$, and let $h(y)=-\exp(y)+3y$. Then, $h(0)<0$, $h(1)>0$, and $h'(y)=-\exp(y)+3>0$. All this implies that $f'(y)-g'(y)$ has a single root on the interval $(0,1)$, which in turn means that $f(y)-g(y)<0$ on the same interval. Therefore, $$\prod\left(x-a_i\right)\le x^n\left(1-\frac{1}{x^6}\right)\le x^n-x^{n-6}\le x^n-1$$ for $n\ge 6$. As @bgins said earlier, it is easy to show that the above holds for $n=2$ and $3$. However, it seems very tedious to show this for $n=4$ and $5$.

POSSIBLE SOLUTION FOR $n=4$: Assume again $a_1=1$, and let $a_2=a$,$a_3=b$,$a_4=c$. The case where all of these are positive is easy to do by induction, so assume at least $c$ is negative. Then, $$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ac)x-abc=\\x^3+\frac{a+b+c}{2}(a+b+c-2x)x-\frac{a^2+b^2+c^2}{2}x-abc.$$ Under the assumption that $c<0$, we have $-1\le a+b+c\le 2<2x$, and it follows that $$\frac{a+b+c}{2}(a+b+c-2x)x\le \frac{-1}{2}(-1-2x)x=x^2+x/2.$$ Therefore, $$(x-a)(x-b)(x-c)\le x^3+x^2+x+1$$ and this proves it.

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Looks like you might be able to get it by induction on $n$. –  bgins May 12 '12 at 22:45
It's clear how to do the induction for positive $a_i$'s, but I don't see it working otherwise. –  Ivan May 12 '12 at 23:21
Well $n=1$ is trivial and $n=2$ is easy, but I have to admit, I don't see how to do the general case, even with Newton's identity. Perhaps using the inductive step on a shift of $x$? –  bgins May 13 '12 at 15:05

Part a) is indeed simple. Let me give a proof for part b). For $k=1$ our condition gives $a_1\ge -\sum_{k=2}^{n}a_k=S.$ Now we apply AM-GM to estimate $$\prod_{k=2}^{n}(x-a_k)\le \left(\frac{(n-1)x-S}{n-1}\right)^{n-1}\le \left(x+\frac{a_1}{n-1}\right)^{n-1}.$$ So we are left to show that $$(x-a_1)\left(x+\frac{a_1}{n-1}\right)^{n-1}\le x^n-a_1^n.$$ Binomial theorem now implies the result, since $$\left(x+\frac{a_1}{n-1}\right)^{n-1}=x^{n-1}+\sum_{k = 2}^{n-2}{\binom{n-1}{k}x^{n-1-k}\left(\frac{a_1}{n-1}\right)^k}+a_1^{n-1}\le \sum_{k=0}^{n-1}x^{n-1-k}a_1^k$$ and $$\binom{n-1}{k}\cdot\frac{1}{(n-1)^k}< 1.$$