The following result is used my proof of Cauchy's Bound for real roots.
Lemma 1: abs(c)n = abs(cn)
Proof:
(1) Assume that c is nonnegative
(2) Then, cn is nonnegative
(3) Then, abs(cn) = cn
(4) Since abs(c) = c, it follows that: cn = abs(c)n
(5) Assume that c is negative
(6) We can assume that n is odd
[Otherwise, cn = (-c)n = abs(c)n = abs(cn) ]
(8) abs(cn) = -cn = (-1)n*cn = (-c)n = abs(c)n
QED
Lemma 2:
Let:
ancn = -an-1cn-1 + .... + -a0
Then:
abs(an)*abs(c)n ≤ abs(an-1)*abs(c)n-1 + ... + abs(a0)
Proof:
(1) Using the Triangle Inequality (see Lemma 4, here), we know that:
abs(-an-1cn-1 + .... + -a0) ≤ abs(-an-1cn-1) + ... + abs(-a0)
so that:
abs(ancn) ≤ abs(-an-1cn-1) + ... + abs(-a0)
(2) Using a basic property of inequalities (see Lemma 1, here):
abs(an)*abs(cn) = abs(ancn)
and likewise:
abs(-an-1)*abs(cn-1) = abs(-an-1cn-1)
...
(3) So we have:
abs(an)*abs(cn) ≤ abs(-an-1)*abs(cn-1) + ... + abs(-a0)
(4) Using Lemma 1 above, we have:
abs(an)*abs(c)n ≤ abs(an-1)*abs(c)n-1 + ... + abs(a0)
QED
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