Let $f(t) = t^n + a_{n-1}t^{n-1} + \cdots + a_0$. Let $\alpha$ be a root of $f$. Then show that $\alpha \leq n \max_{i} |a_i|$.
I could only figure it out for the special case where $|a_i| < 1$ for all $i$. I thought I'd be able to generalize this but I can't figure it out at all. Can someone help me with this?
The result is not quite true. We first prove a correct version, and then show that there are polynomials for which the result does not hold.
Let $M$ be the maximum of the $|a_i|$. We have $$\alpha^n=-\left(a_{n-1}\alpha^{n-1}+\alpha^{n-2}+\cdots +a_0\right).$$ By the Triangle Inequality, it follows that $$|\alpha|^n\le M\left(|\alpha|^{n-1}+|\alpha|^{n-2}+\cdots+1\right).\tag{1}$$ If $|\alpha|\ge 1$ then the left side of (1) is $\le Mn|\alpha|^{n-1}$, and the result follows for this case.
A counterexample: If $|\alpha|\lt 1$, then we can run into trouble. For example, consider the equation $x^2-\frac{x}{4}-\frac{1}{4}=0$. One root is $\frac{\frac{1}{4}+\frac{1}{4}\sqrt{17}}{2}$. The maximum of the $|a_i|$ is $\frac{1}{4}$ and $n=2$, but the root is greater than $\frac{1}{2}$.
One can get a correct result by replacing $\max(|a_i|)$ by $\max(1/n,|a_i|)$, and undoubtedly in many other ways.