If all the coeffcients $b_1,b_2,..., b_{n+1}$ of the polynomial $f(x)$ are negative then $0>b_1=-a_0+a_1+a_2+...+a_n$.

Let $g(x)=x^n+s_1x^{n-1}+...+s_n$. As $a_1,...,a_n$ are positive real numbers, so are $s_1,...,s_n$. $f(x)=x^{n+1}+(s_1-a_0)x^n+(s_2-a_0s_1)x^{n-1}+...+(s_n-a_0s_{n-1})x-a_0s_n$ If $a_0>a_1 + a_2 +...+ a_n=s_1>0$, then we have to prove for all $k=1,2,...,n-1$ that $s_{k+1}<s_1s_k<a_0s_k$ as $-a_0s_n<0$ obviously. But $s_{k+1}<s_1s_k$ is also obvious as all of the terms in the sum $s_{k+1}$ appear in sum after opening brackets in $s_1s_k$ which also have some positive terms.