Suppose we have $a,b,c,d\in {\bf R}[x]$ such that $a+b+c,a+b+d,a+c+d,b+c+d$ have roots and $a+b,a+c,a+d,b+c,b+d,c+d$ have not. To each of the polynomials $a+b,\cdots,c+d$ we can attribute a sign, indicating if they are always positive or always negative. From the triplet $a+b,a+c,a+d$ two have the same sign and by permuting and flipping ($p\rightarrow -p$) we may assume that $a+b,a+c>0$ on ${\bf R}$. If $b+c>0$ then also $(a+b)+(a+c)+(b+c)>0$, contradiction. Hence $b+c<0$ and from $b+c<0<a+b,a+c$ we deduce $b,c<a$. Since $b+c<0$ we have $b+d>0 \vee c+d>0$ (otherwise $b+c+d<0$) and since $a>b,c$ we have $a+d>0$. Choose $x$ such that $a(x)+b(x)+d(x)=0$. Since $a+b,a+d>0$ we have $b(x),d(x)<0$ hence $b+d<0$. Choose $y$ such that $a(y)+c(y)+d(y)=0$. Since $a+c,a+d>0$ we have $d(y),c(y)<0$ hence $c+d<0$. Combining we find $b+c,b+d,c+d<0$ and $(b+c)+(b+d)+(c+d)<0$, contradiction. It follows that such $a,b,c,d$ do not exist. Note that it doesn't matter that $a,b,c,d$ are polynomials, any quadruplet of continuous functions will do.