, 2014

Note first that $b_1 = (a_1 + a_2 + \dots + a_n) - a_0$, so the condition
$$a_0 > a_1 + a_2 + \dots + a_n$$
is necessary for $b_1 < 0$.
Below it will be useful to know that in the case $a_0 = a_1 + a_2 + \dots + a_n$, the coefficients $b_i$ are all non-positive. First note that $a_0 = a_1 + a_2 + \dots + a_n$ implies $b_1 = 0$. Next, write
$$g(x) = (x + a_1) \cdots (x + a_n) = x^n + c_1 x^{n-1} + \dots + c_{n-1} x + c_n.$$
Then, the coefficient $c_j$ is the sum of all the products $a_{r_1}a_{r_2}\cdots a_{r_j}$ with $1 \le r_1 < r_2 < \dots < r_j \le n$. In particular, $c_j > 0$ for $1 \le j \le n$. Moreover, $b_{j+1} = c_{j+1} - a_0c_j$ for $1 \le j \le n-1$ and $b_{n+1} = -a_0c_n < 0$. Observe that when $j < n$ and
$$a_0 c_j = (a_1 + a_2 + \dots + a_n) c_j$$
is multiplied out in full, every term $a_{s_1}s_{s_2}\cdots a_{s_{j+1}}$ with $1 \le s_1 < s_2 < \dots < s_{j+1} \le n$ occurs, since $a_{s_1}s_{s_2}\cdots a_{s_j}$ occurs in $c_j$ and $a_{s_{j+1}}$ in $(a_1+a_2+\dots+a_n)$. Since all $a_i > 0$, we deduce that $(a_1+a_2+\dots+a_n)c_j \ge c_{j+1}$, for $1 \le j \le n-1$. Hence, if $a_0 = a_1 + a_2 + \dots + a_n$, the coefficients $b_i$ are all non-positive. Finally, suppose $a_0 > a_1 + a_2 + \dots + a_n$, and let $v = a_0 - (a_1 + a_2 + \dots + a_n)$. Then $f(x) = (x-(a_1+a_2+\dots+a_n))g(x)-vg(x)$. Since $v > 0$, all coefficients of $-vg(x)$ are negative, and, excepting the coefficient 1 of $x^{n+1}$, all coefficients of $(x-(a_1+a_2+\dots+a_n))g(x)$ are non-positive, as shown above. So all coefficients $b_j$ of $f(x)$ are negative, as required.