10. Find the minimum value of $(a+b)\left(\frac{1}{a}+\frac{4}{b}\right)$, where $a$ and $b$ range over all positive real numbers.
(A) 3
(B) 6
(C) 9
(D) 12
(E) 15
10. Answer: (C) $(a+b)\left(\frac{1}{a}+\frac{4}{b}\right)=1+4+\frac{b}{a}+\frac{4 a}{b}=5+\frac{b^{2}+4 a^{2}}{a b}$. Let $m=\frac{b^{2}+4 a^{2}}{a b}$. Then we have $4 a^{2}-m b a+b^{2}=0$. This implies that for any positive values of $m$ and $b$, $m^{2} b^{2}-16 b^{2} \geq 0$. Thus we have $m^{2}-16 \geq 0$, or $m \geq 4$.
Therefore $(a+b)\left(\frac{1}{a}+\frac{4}{b}\right)=5+m \geq 9$. Hence the minimum value is 9 .