(Problem)
$f(x)=\sum\limits_{k=0}^n a_kx^k$ be a polynomial of real coefficients which satisfies $a_0=1$ and $deg(f(x))=n$. Show that there exists a complex number $\alpha$ satisfying $|f(\alpha)|\geq1 \land |\alpha|=1$.
(My Solution 1)
(Lemma$1$: Lagrange interpolation)
For any polynomial which satisfies $deg(f(x))=n,(n\geq1)$ and $n + 1$ distinct complex numbers $\alpha_0,\alpha_1,..., \alpha_n$
$f(x)=\sum\limits_{k=1}^n f(\alpha_k)\prod\limits_{\substack{0\leq r\leq n\\r\neq k}}\frac{x-\alpha_r}{\alpha_k-\alpha_r}$
is valid.
(Lemma$2$)
When $n$ is a positive integer and $f(x)$ is a monic polynomial of real coefficients which satisfies $deg(f(x))=n$, for any $n+1$ distinct complex numbers $\alpha_0,\alpha_1,..., \alpha_n$
$\sum\limits_{k=0}^n \frac{f(\alpha_k)}{\prod\limits_{\substack{0\leq r\leq n\\r\neq k}}(\alpha_r-\alpha_k)}\geq1$
is valid.
(This can be shown by using Lemma1 and the triangle inequality)
When $\alpha_0,\alpha_1,..., \alpha_n$ is a solution of $x^{n+1}-1=0$ , $|\alpha_k|=1$.
$(n+1)x^{n}=\prod\limits_{\substack{0\leq r\leq n\\r\neq k}}(x-\alpha_r)$
because $x^{n+1}-1=\prod\limits_{k=0}^n (x-\alpha_k)$
From this
$(n+1)\alpha_k^{n}=\prod\limits_{\substack{0\leq r\leq n\\r\neq k}}(\alpha_k-\alpha_r)$
Which means
$\sum\limits_{k=0}^n \frac{1}{\prod\limits_{\substack{0\leq r\leq n\\r\neq k}}|\alpha_r-\alpha_k|}=\frac{1}{n+1} \sum\limits_{k=0}^n \frac{1}{|\alpha_k|^n}=1$.
From this and Lemma$2$,an $\alpha_k$ which satisfies $|f(\alpha_k)|\geq1$ exists.
This is my first solution using Lagrange interpolation. Also I found a solution using complex analysis. In this way, I used the fact that generally when $f(x)=\sum\limits_{k=0}^n a_kx^k$
$a_k=\frac{1}{2\pi i}\int_{C}^{}\frac{f(z)}{z^{n+1}}$ is valid.
This second solution using complex analysis can show a general version of the Problem.(I will call this Problem2)
(Problem2) Let $f(x)=\sum\limits_{k=0}^n a_kx^k$ be a polynomial of real coefficients which satisfies $max a_k\geq1, (0\leq k\leq n)$.Show that there exists a complex number $\alpha$ satisfying $|f(\alpha)|\geq1 \land |\alpha|=1$.
(My Question)
Problem2 can be shown by using complex analysis. However I want to know a elementary solution(a solution without using complex analysis) like the one I shown as (My solution1).
Can anyone solve (Problem2) in a elementary way?
