Sorry for reposting this, it is of course very well-known. It was pretty much impossible to search for it.. all I could find was http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=283732 Anyway, the standard proof goes like $LHS=(a+c)\left(\frac{1}{a+b}+\frac{1}{c+d}\right)+(b+d)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)$. $\frac{1}{a+b}+\frac{1}{c+d}\ge\frac{4}{a+b+c+d}$ and $\frac{1}{b+c}+\frac{1}{d+a}\ge\frac{4}{a+b+c+d}$ and the result follows.

By Titu inequality, $\sum_{cyc}\frac{a+c}{a+b}=\sum_{cyc}\frac{(a+c)^2}{(a+b)(a+c)}\geq \frac{4(a+b+c+d)^2}{(a+b+c+d)^2}=4$