$13 \cdot 24$ If $a$ and $b$ are positive real numbers, and the equations $x^{2}+a x+2 b=0$ and $x^{2}+2 b x+a=0$ both have real roots, then the smallest positive value of $a+b$ is
(A) 2 .
(B) 3 .
(C) 4 .
(D) 5 .
(E) 6 .
(35th American High School Mathematics Examination, 1984)
[Solution] Given that the two equations provided in the problem have real roots, the discriminants are non-negative. Therefore, we have the following inequalities: $a^{2}-8 b \geqslant 0$ and $4 b^{2}-4 a \geqslant 0$, i.e.,
$$
a^{2} \geqslant 8 b \text { and } \quad 4 b^{2} \geqslant 4 a \text {. }
$$

Thus,
$$
a^{4} \geqslant 64 b^{2} \geqslant 64 a \text {. }
$$

Since $a>0$, we get $a^{3} \geqslant 64$, which means $a \geqslant 4$.
Also, since $b^{2} \geqslant a$, we get $b \geqslant 2$.
Moreover, $a=4$ and $b=2$ also make both equations have real roots. Therefore, the minimum value of $a+b$ is 6. Hence, the answer is $(E)$.