smc

Since both of the equations have real roots, both of their discriminants are nonnegative. Therefore, we have $a^2-4(1)(2b)=a^2-8b\ge 0⟹a^2\ge 8b$ from the first equation, and $(2b{)}^2-4(1)(a)=4b^2-4a\ge 0⟹4b^2\ge 4a⟹b^2\ge a$ from the second. We can square the second equation to get $b^4\ge a^2$, and combining this with the first one gives $b^4\ge a^2\ge 8b$, so $b^4\ge 8b$. We can divide both sides by $b$, since it is positive, and take the cubed root of that to get $b\ge 2$. Therefore, we have $a^2\ge 8b\ge 8(2)=16$, and since $a$ is positive, we can take the square root of this to get $a\ge 4$. Therefore, $a+b\ge 2+4=6$, and the smallest possible value is $6,\boxed{6}$.