Let $P(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_{1}x + 1$ where $a_i$ are nonnegative and real. Assume $P$ has $n$ real roots.
Prove $P(2) \geq 3^n$.
I thought I had a good idea about rewriting $P$ as $(x-\alpha_1)\cdots(x-\alpha_n)$. The fact that the roots are real means that you can order them. Choose the largest root, $\alpha_k$, then $2-\alpha_k$ would have the smallest such value among the $\alpha_i$. And so we would have $P(2) \geq (2-\alpha_k)^n$.
But I haven't been able to think of a way to compare it with $3^n$.
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