jmc

Note that $x=-1$ is always a root of $x^3+a{x}^2+ax+1=0,$ so we can factor out $x+1,$ to get $$(x+1)(x^2+(a-1)x+1)=0.$$The quadratic factor has real roots if and only if its discriminant is nonnegative: $$(a-1{)}^2-4\ge 0.$$This reduces to $a^2-2a-3\ge 0,$ which factors as $(a+1)(a-3)\ge 0.$ The smallest positive value that satisfies this inequality is $\boxed{3}.$