TITLE: Polynomial puzzle
QUESTION [11 upvotes]: A nice problem that I thought I would share with you math hobbyists:

Let $P$ be a polynomial of even degree $n$ with real coefficients and $n$ real positive roots, defined by
  $$P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_2x^2-nx+a_0$$
  such that $P(0)\ge 1$. Prove that $P(2)=1$.

PS: I already know a solution. I'm just sharing this for fun.

REPLY [5 votes]: Let $ x_{1}, \dots , x_{n} $ be the positive real roots of $ P $. We have that $$ \prod_{i=1}^{n}x_{i}=(-1)^{n}a_{0}=a_{0} $$ and $$ \sum_{i=1}^{n}\prod_{j \neq i}x_{j}=(-1)^{n-1} \cdot (-n)=n $$
The last equality can be rewritten as $$ \prod_{i=1}^{n}x_{i} \cdot \sum_{i=1}^{n} \frac{1}{x_{i}}=n $$
Therefore the harmonic mean of the $ x_{i}'s $ is $$ \frac{n}{\sum_{i=1}^{n} \frac{1}{x_{i}}}=\prod_{i=1}^{n}x_{i} $$
On the other hand, we know that $$ \frac{n}{\sum_{i=1}^{n} \frac{1}{x_{i}}} \leq (\prod_{i=1}^{n}x_{i})^{\frac{1}{n}} \leq \prod_{i=1}^{n}x_{i} $$ since  $$ \prod_{i=1}^{n}x_{i}=a_{0}=P(0) \geq 1 $$
So the harmonic mean and the geometric mean of the $ x_{i}'s$ must be equal, thus the $ x_{i}'s $ must be equal to some positive real number $ a $ and since $$ a^{n}=\frac{n}{\frac{n}{a}} $$ we conclude that $ a=1 $ is the unique root of $ P $ of multiplicity $ n $. So $ P(x)=(x-1)^{n} $ which implies that $ P(2)=1 $.