12) Let 4 positive integers $a, b, c, d$ be fixed such that $\frac{a}{b} \leq \frac{c}{d} \leq 1$. Which of the following inequalities is certainly true?
(A) $\frac{a+c}{b+d} \geq \frac{a}{b}+\frac{c}{d}$
(B) $\frac{a+c}{b+d} > 1$
(C) $\frac{a+c}{b+d} \leq \frac{c}{d}$
(D) $\frac{a+c}{b+d} < 1$
(E) none of the above
12. The answer is $(\mathrm{C})$.

Indeed (since $b>0$ and $d>0$) the inequality $\mathbf{C}$ is equivalent to $d(a+c) \leq c(b+d)$, which in turn is equivalent to $a d \leq b c$, or $\frac{a}{b} \leq \frac{c}{d}$. The other answers are all incorrect. Taking $a=b=c=d=1$, one would get $1 \geq 2$ from $\mathbf{A}$, and $1>1$ from $\mathbf{D}$, which are clearly false inequalities. Taking instead $a=c=1$ and $b=d=2$, $\mathbf{B}$ would yield $\frac{1}{2} \geq 1$, which is absurd. [Problem proposed by P. Leonetti.]