abstract algebra - 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 $3^n$.

Answer

The proof follows from the following Lemma:

Lemma If $0

$$(2+a)(2+b) \geq 3(2+ab)$$

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

$$(2+b_1)(2+b_2)(2+b_3)\cdot ... \cdot (2+b_n) \geq 3^n \sqrt[3]{b_1b_2...b_n}=3^n$$