Polynomial puzzle

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 $.

Since $P$ has $n$ real roots we can write \begin{eqnarray} P(x) & = & x^n+a_{n-1}x^{n-1}+\cdots a_2x^2 -nx + a_0 \\ & = & (x-x_1)(x-x_2)\cdots(x-x_n) \\ & = & x^n + a_{n-1}x^{n-1} + \cdots a_2x^2 +(-1)^{n-1}(\sum_{i=1}^n\prod_{j\neq i} x_j)\, x + x_1x_2\cdots x_n \end{eqnarray} Therefore $x_1x_2\cdots x_n = a_0$ and $(-1)^{n-1}(\sum_{i=1}^n\prod_{j\neq i} x_j) = -n$. But since $n$ is even the latter equality together with $\prod_{j\neq i}x_j = \dfrac{a_0}{x_i}$ gives $\sum_{i=1}^n\dfrac{a_0}{x_i} = n$.

By assumption $P(0)= a_0 = x_1x_2\cdots x_n\geq1$. However, by the AM-GM inequality, $$1=\frac{\sum_{i=1}^n\dfrac{a_0}{x_i}}{n}\geq a_0\left(\frac{1}{x_1x_2\cdots x_n}\right)^\frac{1}{n} = a_0^{1-\frac{1}{n}}\geq 1$$ with equality only at $\dfrac{a_0}{x_1}=\dfrac{a_0}{x_2}=\ldots=\dfrac{a_0}{x_n}$. Because equality holds, we know that all $\dfrac{a_0}{x_i}=1$ and therefore $\dfrac{a_0^n}{x_1x_2\cdots x_n} = a_0^{n-1} = 1$ which implies $a_0=1$ and hence $x_i=1.$ $P(2) = 1$ now follows directly as desired.