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Let $p(x)=a_0x^n+a_1x^{n-1}+\cdots+a_n,a_0 \ne 0$ to be univariate (1 variable) polynomial of degree $n$.

Let $r$ be its root, i.e. $p(r)=0$.

How can I prove that: $$|r| \le \max\left(1, \sum_{i=1}^n \left|{a_i \over a_0}\right|\right)$$

And is this always true? $$|r| \le \sum_{i=1}^n \left|{a_i \over a_0}\right|$$

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Hint: proof by contradiction. If $|z| > max(1,\sum_{i=1}^n |\frac{a_i}{a_0}|)$, verify the leading term $|a_0 z^n|$ is bigger than the rest of terms. –  achille hui Feb 5 '13 at 17:28
    
As for your second question, the sum $\sum_{i=1}^n |{a_i \over a_0}|$ will always be greater than or equal to $1$, since $|\frac{a_0}{a_0}| = 1$. Writing $\max(1, \sum_{i=1}^n |{a_i \over a_0}|)$ is thus kind of redundant (you might see it as a hint). So if your original statement is true, then your last statement is also true. –  Arthur Feb 5 '13 at 17:29

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Hint: Suppose that $|r|\ge 1$. Then for any $k$, $|a_kr^k|\le |a_kr^{n-1}|$.

It follows (Triangle Inequality) that $$|a_1r^{n-1}+a_2r^{n-2}+\cdots +a_n|\le |r^{n-1}|(|a_1|+|a_2|+\cdots +|a_n|).$$

But if $a_0r^n+a_1r^{n-1}+\cdots+a_n=0$, then $|a_0r_n|=|a_1r^{n-1}+a_2r^{n-2}+\cdots +a_n|$.

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@BluePonyInc.: Please note there was a bad typo/mistake in the first version you saw. –  André Nicolas Feb 5 '13 at 17:44
    
Thanks, fixed it on paper just now. –  Blue Pony Inc. Feb 5 '13 at 17:47

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