If $a$ and $b$ are positive [real numbers](https://artofproblemsolving.com/wiki/index.php/Real_numbers) and each of the [equations](https://artofproblemsolving.com/wiki/index.php/Equation) $x^2+ax+2b=0$ and $x^2+2bx+a=0$ has real [roots](https://artofproblemsolving.com/wiki/index.php/Root_(polynomial)), then the smallest possible value of $a+b$ is
$\mathrm{(A) \ }2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ } 6$
Since both of the equations have real roots, both of their [discriminants](https://artofproblemsolving.com/wiki/index.php/Discriminant) are nonnegative. Therefore, we have 
$a^2-4(1)(2b)=a^2-8b\geq0\implies a^2\geq8b$ from the first equation, and 
$(2b)^2-4(1)(a)=4b^2-4a\geq0\implies 4b^2\geq4a\implies b^2\geq a$ from the second.
We can square the second equation to get $b^4\geq a^2$, and combining this with the first one gives $b^4\geq a^2\geq8b$, so $b^4\geq8b$. We can divide both sides by $b$, since it is positive, and take the cubed root of that to get $b\geq2$. Therefore, we have $a^2\geq8b\geq8(2)=16$, and since $a$ is positive, we can take the [square root](https://artofproblemsolving.com/wiki/index.php/Square_root) of this to get $a\geq4$. Therefore, $a+b\geq 2+4=6$, and the smallest possible value is $6, \boxed{\text{E}}$.