Lemma 1:Let $ x_1,x_2,..,x_n$ are positive real numbers such that $ x_i\leq 1$ for all $ i = 1,2..,n$(or $ x_i\geq 1$ for all $ i = 1,2..,n$ .We have: $ x_1 + x_2 + .. + x_n\leq n - 1 + x_1x_2...x_n$ Lemma 2:Let polynomial $ f(x) = g(x)h(x)$($ f,g,h\in C[x]$) that: $ f(x) = a_0x^n + a_1x^{n - 1} + ... + a_{n - 1}x + a_n$ $ g(x) = b_0x^k + b_1x^{k - 1} + ... + b_{n - 1}x + b_n$ $ h(x) = c_0x^l + c_1x^{l - 1} + ... + c_{n - 1}x + c_n$ We have:$ |b_0c_l|^2 + |c_0b_k|^2\leq |a_0|^2 + |a_1|^2 + ... + |a_n|^2$ Let $ z_1,z_2,..,z_n$ are root of polynomial $ f(x)$;$ z_1,z_2,..,z_k$ are root of polynomial $ f(x)$ such that :$ |z_i|\leq 1$ for all $ i = 1,2,..,k$;$ z_{k + 1},..,z_n$ are root of polynomial $ f(x)$ such that:$ |z_i| > 1$ for all $ i = k + 1,..,n$ Using lemma 1:we have:$ \sum_{i = 1}^{n}|z_i|^2\leq k - 1 + n - k - 1 + |z_1..z_k|^2 + |z_{k + 1}..z_n|^2$. Using lemma 2 for $ g(x) = (x - z_1)..(x - z_k)$ and $ h(x) = (x - z_{k + 1})..(x - z_n)$ we have:$ |z_1z_2...z_k|^2 + |z_{k + 1}..z_n|^2\leq 1 + |a_1|^2 + ... + |a_n|^2\leq 2$. This problem is proven

Very Nice Solution~ this seems dumb but could you provide a proof of lemma 2?

The key is to notice $f(x)\overline{f}(1/x)$ gives the sum of squares of coefficient magnitudes, so we can "swap" roots with small magnitude to get the desired bound. This also gives an upper bound on the Mahler measure.

I am sorry, but could someone clarify what math154 means by swapping roots with small magnitude? Thank you very much!

Let's begin with the first part: "the key is to notice $f(x)\overline{f}(1/x)$ gives the sum of squares of coefficient magnitudes". It should mean: $$ (*)\,\,\,\,\,\,\,\,\,\frac{1}{2\pi i}\oint_C \frac{1}{z}f(z)\overline{f}(\frac{1}{z})\,dz=\sum_{k = 1}^n |a_k|^2. $$where $C$ is the unit circle. Indeed, it follows from: \[\frac{1}{2\pi i}\oint_C z^n\,dz =\begin{cases}1 \,,\,n=-1\\ 0\,,\, n\neq -1, n\in \mathbb{Z}\end{cases}\]See that the LHS of (*) equals $\frac{1}{2\pi}\oint_C |f(z)|^2\,|dz|=\frac{1}{2\pi}\int_0^{2\pi}|f(e^{i\varphi})|^2 \,d\varphi$. Now, using AM-GM, we get: \[\frac{1}{2\pi}\int_0^{2\pi}|f(e^{i\varphi})|^2 \,d\varphi\geq \exp \left(2\left(\frac{1}{2\pi}\int_0^{2\pi}\ln\left(|f(e^{i\pi}|)\right)\,d\varphi \right)\right) =\left(M(f)\right)^2 \]where $M(f)$ is the Mahler measure of the polynomial $f$. Let $z_1,z_2,\dots,z_m$ be the zeroes of $f$ with magnitude greater or equal to $1$. We have (using Bernouli inequality): \[\left(M(f)\right)^2=\prod_{i=1}^m |z_i|^2\geq 1+\sum_{i=1}^m (|z_i|^2-1)=\sum_{i=1}^m |z_i|^2-m+1\]All that together yields: \[\sum_{i=1}^m |z_i|^2\leq m.\]Summing up the rest $n-m$ of $z_i$'s with small magnitude, the result follows.

Cool, thank you so much for all your help, dgrozev! I don't mean to be picky, since your solution is truly amazing, but I was wondering if there is a solution without calculus?

MathPanda1 wrote: ...but I was wondering if there is a solution without calculus? yes, there is, the one in the post #2. But what you asked was about the meaning behind this: math154 wrote: The key is to notice $f(x)\overline{f}(1/x)$ gives the sum of squares of coefficient magnitudes,... and I cannot think of something that doesn't it involve some complex analysis.

Thank you very much for all your help! I was just wondering, since I wasn't able to prove the second lemma in math10's solution. I was wondering if you have a proof of that lemma? Thank you!

dgrozev wrote: Let's begin with the first part: "the key is to notice $f(x)\overline{f}(1/x)$ gives the sum of squares of coefficient magnitudes". It should mean: $$ (*)\,\,\,\,\,\,\,\,\,\frac{1}{2\pi i}\oint_C \frac{1}{z}f(z)\overline{f}(\frac{1}{z})\,dz=\sum_{k = 1}^n |a_k|^2. $$where $C$ is the unit circle. This is cool. However, the right hand side is the sum of ALL normed squares of coefficients of $f$, including the leading coefficient $1$. So we can only get from (*) that $|z_1|^2 + \dots + |z_m|^2 \leq m+1$.