Solutions

Let

$$f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0.$$

For all $x$ with

$$|x|\ge M=\max(1,2n|a_{n-1}|,\ldots,2n|a_0|),$$

we have

$$\frac12\le 1+\frac{a_{n-1}}x+\cdots+\frac{a_0}{x^n}\le \left|1+\frac{a_{n-1}}x+\cdots+\frac{a_0}{x^n}\right|.$$

Hence, for $|x|\ge M$,

$$\left|\frac{x^n}{2}\right|\le |x^n|\left|1+\frac{a_{n-1}}x+\cdots+\frac{a_0}{x^n}\right|=|f(x)|.$$

Now, let $b>M$ be a number such that

$$\frac{|b^n|}{2}\ge |f(0)|.$$

It follows that for $|x|\ge b$,

$$|f(x)|\ge\frac{|x^n|}{2}\ge\frac{|b^n|}{2}\ge |f(0)|.$$

Since $f(x)$ is continuous, so is $|f(x)|$, and thus there exists $y\in[-b,b]$ such that

$$|f(y)|\le |f(x)|$$

for all $x$ in $[-b,b]$. We conclude that $|f(y)|$ is the global minimum of $|f|$ because $|f(y)|\le |f(0)|$.