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I do not see clearly how to use the information that the coefficients be integer numbers, apart the fact that $1\le |b_m|\le |a_n|$; therefore the following estimate is possibly non-optimal. However, it gives a simple and explicit bound, that also holds for complex coefficients.

By a classic estimate on the zeros of a polynomial, all (complex) roots of $f$ are bounded in modulus by $$1+\max_{0\le i < n}\frac{|a_i|}{|a_n|}\le M+1\, ,$$ because $|a_m|\ge1$. Since $b_i$ is the $i$-th elementary symmetric polynomial of some $m$ of the roots of $f$, times $b_m$, which is less that $M$ is size, we have

$$|b_i|\le M {m \choose i}(M+1)^i\, ,$$ so for instance $$\sum_{i=0}^n|b_i|\le M':=M(M+2)^n\, .$$

I do not see clearly how to use the information that the coefficients be integer numbers, apart the fact that $1\le |b_m|\le |a_n|$; therefore the following estimate is possibly non-optimal. However, it gives a simple and explicit bound, that also holds for complex coefficients.

By a classic estimate on the zeros of a polynomial, all (complex) roots of $f$ are bounded in modulus by $$1+\max_{0\le i < n}\frac{|a_i|}{|a_n|}\le M+1\, ,$$ because $|a_m|\ge1$. Since $b_i$ is the $i$-th elementary symmetric polynomial of some $m$ of the roots of $f$, times $b_m$, which is less that $M$ is size, we have

$$|b_i|\le M {m \choose i}(M+1)^i\, ,$$ so for instance $$\sum_{i=0}^m|b_i|\le M':=M(M+2)^m\, .$$

I do not see clearly how to use the information that the coefficients be integer numbers, apart the fact that $1\le |b_m|\le |a_n|$; therefore the following estimate is possibly non-optimal. However, it gives a simple and explicit bound, that also holds for complex coefficients.

By a classic estimate on the zeros of a polynomial, all (complex) roots of $f$ are bounded in modulus by $$1+\max_{0\le i < n}\frac{|a_i|}{|a_n|}\le M+1\\ ,$$ because $|a_m|\ge1$. Since $b_i$ is the $i$-th elementary symmetric polynomial of some $m$ of the roots of $f$, times $b_m$, which is less that $M$ is size, we have

$$|b_i|\le M {m \choose i}(M+1)^i\\ ,$$ so for instance $$\sum_{i=0}^n|b_i|\le M':=M(M+2)^n\\ .$$

I do not see clearly how to use the information that the coefficients be integer numbers, apart the fact that $1\le |b_m|\le |a_n|$; therefore the following estimate is possibly non-optimal. However, it gives a simple and explicit bound, that also holds for complex coefficients.

By a classic estimate on the zeros of a polynomial, all (complex) roots of $f$ are bounded in modulus by $$1+\max_{0\le i < n}\frac{|a_i|}{|a_n|}\le M+1\, ,$$ because $|a_m|\ge1$. Since $b_i$ is the $i$-th elementary symmetric polynomial of some $m$ of the roots of $f$, times $b_m$, which is less that $M$ is size, we have

$$|b_i|\le M {m \choose i}(M+1)^i\, ,$$ so for instance $$\sum_{i=0}^n|b_i|\le M':=M(M+2)^n\, .$$

I do not see clearly how to use the information that the coefficients be integer numbers, apart the fact that $1\le |b_m|\le |a_n|$. The following is an explicit bound, that also holds for complex coefficients.

By a classic estimate on the zeros of a polynomial, all (complex) roots of $f$ are bounded in modulus by $$1+\max_{0\le i < n}\frac{|a_i|}{|a_n|}\le M+1\\ ,$$ because $|a_m|\ge1$. Since $b_i$ is the $i$-th elementary symmetric polynomial of some $m$ of the roots of $f$, times $b_m$, which is less that $M$ is size, we have

$$|b_i|\le M {m \choose i}(M+1)^i\\ .$$

I do not see clearly how to use the information that the coefficients be integer numbers, apart the fact that $1\le |b_m|\le |a_n|$; therefore the following estimate is possibly non-optimal. However, it gives a simple and explicit bound, that also holds for complex coefficients.

By a classic estimate on the zeros of a polynomial, all (complex) roots of $f$ are bounded in modulus by $$1+\max_{0\le i < n}\frac{|a_i|}{|a_n|}\le M+1\\ ,$$ because $|a_m|\ge1$. Since $b_i$ is the $i$-th elementary symmetric polynomial of some $m$ of the roots of $f$, times $b_m$, which is less that $M$ is size, we have

$$|b_i|\le M {m \choose i}(M+1)^i\\ ,$$ so for instance $$\sum_{i=0}^n|b_i|\le M':=M(M+2)^n\\ .$$