We wish to show that the inequality holds, with equality only when $a/b=c/d$
Consider the function $f(t)=(c/d)t+(a/b)(1-t)$. This is a linear function equal to $a/b$ when $t=0$ and $c/d$ when $t=1$. Moreover, $f(t)$ is between these two quantities if and only if $0\leq t\leq 1$. Therefore, if we can show that $f(t)=(a+c)/(b+d)$ for some $t$ between $0$ and $1$, we will be done.
Let us solve:
Multiplying by $b*d$, expanding out, and rearranging terms, we get
Assuming that $bc=ad$ is nonzero (which happens when $a/b$ and $c/d$ are not equal), we can divide out to get $t=d/(b+d)$. Assuming that $d>0$ and $b>0$, or that $b<0$ and $d<0$ we have $t$ is between $0$ and $1$, and we are done. However, if the signs of $b$ and $d$ are different, then $t$ will be outside the range, and the inequality will not hold.