Polynomial with n real roots 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 $3^n$. 
 A: The proof follows from the following Lemma:
Lemma If $0<a \leq 1 \leq b$ then 
$$(2+a)(2+b) \geq 3(2+ab)$$
Proof:
$$(2+a)(2+b) \geq 3(2+ab) \Leftrightarrow \\
4+2a+2b+ab \geq 6+3ab \Leftrightarrow \\
0 \geq 2-2a-2b+2ab \Leftrightarrow \\
0 \geq 2(a-1)(b-1)
$$
QED Lemma
Now, lets solve the problem. Let $b_i=-\alpha_1$, and we can assume without loss of generality that $b_1 \leq b_2 \leq ... \leq b_n$.
As $b_1 \cdot .. \cdot b_n=1$ it follows that 
$$b_n \geq 1 \\
b_1\cdot ... \cdot b_k \leq 1$$
Then, by repeadely applying the previous lemma, we get
$$(2+b_1)(2+b_2)(2+b_3)\cdot ... \cdot (2+b_n) \geq \\
3(2+b_1b_2)(2+b_3)\cdot ... \cdot (2+b_n) \geq \\
3^2(2+b_1b_2b_3)\cdot ... \cdot (2+b_n) \geq \\
....\\
3^{k-1}(2+b_1b_2b_3...b_k)(2+b_{k+1}\cdot ... \cdot (2+b_n) \geq \\
...\\
3^{n-1}(2+b_1b_2b_3...b_n)=3^n \\
$$
Much simpler second solution:
If $b_1,..,b_n$ are non-negative numbers then by the AM-GM inequality:
$$2+b_i = 1+1+b_i \geq 3 \sqrt[3]{b_i}$$
Therefore
$$(2+b_1)(2+b_2)(2+b_3)\cdot ... \cdot (2+b_n) \geq 3^n  \sqrt[3]{b_1b_2...b_n}=3^n$$
