Here we go. Since $g(\cdot)$ does not have any real roots, we get $b^2<2a$. Now, let $r_1,r_2,r_3$ be the roots of the cubic $f(x)=x^3+ax^2+2bx-1$. By Vieta's formulas, we have $r_1+r_2+r_3=-a$, $r_1r_2+r_1r_3+r_2r_3=2b$, and $r_1r_2r_3=1$. This implies $r_1^2+r_2^2+r_3^2=a^2-4b$. Now, we also have $|r_1|\cdot |r_2|\cdot |r_3|=1$, and thus by AM-GM, we have $a^2-4b\geqslant 3$. Now suppose the hypothesis is false, and that, $a\leqslant b+1$. Then, on the one hand, $b^2-2b+1\geqslant a^2-4b\geqslant 3$, implying that $b^2\geqslant 2b+2$. On the other hand; $b^2<2a\leqslant 2b+2$. These two together clearly yield a contradiction, and thus, the hypothesis $a-b\leqslant 1$ is false.