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?
 A: By C-S
$$\sum_{cyc}\frac{a+c}{a+b}=(a+c)\left(\frac{1}{a+b}+\frac{1}{c+d}\right)+(b+d)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)\geq$$
$$\geq\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{b+c+d+a}=4.$$
A: $$\frac{\color{blue}{a+c}}{a+b}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{blue}{c+a}}{c+d}+\frac{\color{red}{d+b}}{d+a}=\frac{\color{blue}{a+c}}{a+b}+\frac{\color{blue}{a+c}}{c+d}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{red}{b+d}}{d+a}$$
$$\frac{\color{blue}{a+c}}{a+b}+\frac{\color{blue}{a+c}}{c+d}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{red}{b+d}}{d+a}=(\color{blue}{a+c})\color{green}{\left(\frac{1}{a+b}+\frac{1}{c+d}\right)}+(\color{red}{b+d})\color{brown}{\left(\frac{1}{b+c}+\frac{1}{d+a}\right)}$$
WLOG  $$\color{brown}{\frac{1}{b+c}+\frac{1}{d+a}} \ge \color{green}{\frac{1}{a+b}+\frac{1}{c+d}}  $$ 
Then
$$(\color{blue}{a+c})\color{green}{\left(\frac{1}{a+b}+\frac{1}{c+d}\right)}+(\color{red}{b+d})\color{brown}{\left(\frac{1}{b+c}+\frac{1}{d+a}\right)} \ge (\color{blue}{a+c}+\color{red}{b+d})\color{green}{\left(\frac{1}{a+b}+\frac{1}{c+d}\right)}$$ $$= (\color{turquoise}{a+b}+\color{orange}{c+d})\left(\color{turquoise}{\frac{1}{a+b}}+\color{orange}{\frac{1}{c+d}}\right) \ge (1+1)^2=4$$
