Show that $P(m) \ge (m+1)^n$ for all $m \in \mathbb{Z}^+$

Question

I need help in this problem.

Problem: Let polynomial $P(x) = x^n + a_1 x^{n-1} + \dots + a_{n-1} x + 1 \in \mathbb{R}[x]$, in which $n \ge 1$ and $a_i \ge 0, \forall i = 1,\dots,n-1$. Show that $P(m) \ge (m+1)^n$ for all $m \in \mathbb{Z}^+$.

Answer 1

Well, $P(m) = m^n+a_{n-1}m^{m-1}+\ldots+a_1n+1 \geq m^n+1$ for each nonnegative integer $m$ since all coefficients are nonnegative.

In particular, if $a_1=\ldots=a_{n-1}=0$, then $P(m)=m^n+1$. So the inequality (or bound) is tight.

Answer 2

This statement is false. We could simply set $a_i=0$ for all $i \in \{0,1,2,...,n\}$, and now we'd be claiming:

$$\forall m \in \mathbb{Z}, 0 \geq (m+1)^n$$

which I think you can figure out is not correct