6. $a$ and $b$ are both positive real numbers, the equations $x^{2}+a x+2 b=0$ and $x^{2}+2 b x+a=0$ both have real roots, the minimum value of $a+b$ is $\qquad$ .
6. 6 .
$\Delta_{1}=a^{2}-8 b \geqslant 0$, i.e., $a^{2} \geqslant 8 b ; \Delta_{2}=4 b^{2}-4 a \geqslant 0$, i.e., $b^{2} \geqslant a$. Thus $b^{2} \geqslant a \geqslant \sqrt{8 b}$, which gives $b \geqslant$ $2, a \geqslant 4$, so $a+b \geqslant 6$.