Solution:
Let $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$, so $a_{k}>0$ for every $k$. Since the statement holds for $x=1$, $P(1) \geq 1$. Then by Cauchy-Schwarz,
$$P(x) P\left(\frac{1}{x}\right)=\left(\sum_{k=0}^{n}\left(\sqrt{a_{k} x^{k}}\right)^{2}\right)\left(\sum_{k=0}^{n}\left(\sqrt{\frac{a_{k}}{x^{k}}}\right)^{2}\right) \geq\left(\sum_{k=0}^{n} a_{k}\right)^{2}=P(1)^{2} \geq 1$$