Solution: Claim: \(|x_1 + x_2| \leq |x_1| + |x_2|\)
Proof: Case 1: \(x_1, x_2 \geq 0\). The inequality is equivalent to \(|x_1 + x_2| = x_1 + x_2 = |x_1|+|x_2|\) so it's an equality.
Case 2: \(x_1, x_2 \leq 0\). The inequality is equivalent to \(|x_1+x_2| = -x_1-x_2 = |x_1|+|x_2\), so it's also an equality in this case.
Case 3: (wlog) \(|x_1| \geq |x_2| > 0\) and \(x_1x_2 < 0\) then
\(|x_1+x_2| = x_1-x_2 \leq x_1 \leq |x_1|+|x_2|\)
We can prove this by induction, we've already proven the base case and:
\(|x_1+x_2 + \cdots + x_n| \leq |x_1 + x_2 + \cdots x_{n-1}| + |x_n| \leq |x_1| + |x_2| + \cdots + |x_n|\)
- \(\,\) \begin{align*}
&& |f(x) - 1| &= |a_1 x + a_2x^2 + \cdots + a_{n-1}x^{n-1} + x^n| \\
&&&\leq |a_1x| + |a_2x^2| + \cdots + |a_{n-1}x^{n-1}| + |x^n| \\
&&&\leq |a_1||x| + |a_2||x|^2 + \cdots + |a_{n-1}||x|^{n-1} + |x|^n \\
&&&\leq A|x| + A|x|^2 + \cdots + A|x|^{n-1} + |x|^n \\
&&&=A|x| \frac{1-|x|^{n-1}}{1-|x|} + |x|^n \\
&&&= \frac{A|x|-A|x|^{n}+|x|^{n+1}-|x|^n}{1-|x|} \\
&&&= \frac{A|x|-|x|^n(\underbrace{A-|x|+1}_{\geq0})}{1-|x|} \\
&&&\leq \frac{A|x|}{1-|x|}
\end{align*}
- If \(f(\omega) = 0\) then \begin{align*}
&& 1 & \leq \frac{A|\omega|}{1-|\omega|} \\
\Leftrightarrow && 1-|\omega| &\leq A |\omega| \\
\Leftrightarrow && 1 &\leq (1+A) |\omega| \\
\Leftrightarrow && \frac{1}{1+A} &\leq |\omega| \\
\end{align*}
We also know \(\omega \leq 1 < 1 + A\)
- If \(\omega\) is a root of \(f(x)\) then \(1/\omega\) is a root of \(1 + a_{n-1}x + a_{n-2}x^2 + \cdots + a_1x^{n-1}+x^n\) and so \(1/\omega\) satisfies that inequality, ie
\begin{align*}
&& \frac{1}{1+A} && \leq &&|1/\omega| && \leq &&1 + A \\
\Leftrightarrow &&1+A && \geq&& |\omega| && \geq&& \frac{1}{1 + A}
\end{align*}
- First notice that it's equivalent to:
\(0 = x^5 - 1x^4 - \frac{100}{135}x^3-\frac{91}{135}x^2-\frac{126}{135} + 1\)
therefore all integer roots must be between \(-2,-1\) and \(1\) and \(2\).
\(1\) doesn't work. \(-1\) works. Clearly \(2\) cannot work by parity argument, therefore the only integer root is \(-1\).