Prove the inequality $\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a}\geq 4$ $a, b, c, d$ are positive reals.
How would I prove the inequality
$$\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a} \geq 4$$
I have tried using the rearrangement inequality with $a\leq b\leq c\leq d$
But it doesn't seem to work well.
Any hints please?

By C-S
$$\sum_{cyc}\frac{a+c}{a+b}=(a+c)\left(\frac{1}{a+b}+\frac{1}{c+d}\right)+(b+d)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)\geq$$
$$\geq\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{b+c+d+a}=4.$$
