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$a, b, c, d$ 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?

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    $\begingroup$ You could try the hard way, and add all those fractions up and see what you get. $\endgroup$ – Riccardo Orlando May 1 '16 at 6:56
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$$\frac{\color{blue}{a+c}}{a+b}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{blue}{c+a}}{c+d}+\frac{\color{red}{d+b}}{d+a}=\frac{\color{blue}{a+c}}{a+b}+\frac{\color{blue}{a+c}}{c+d}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{red}{b+d}}{d+a}$$

$$\frac{\color{blue}{a+c}}{a+b}+\frac{\color{blue}{a+c}}{c+d}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{red}{b+d}}{d+a}=(\color{blue}{a+c})\color{green}{\left(\frac{1}{a+b}+\frac{1}{c+d}\right)}+(\color{red}{b+d})\color{brown}{\left(\frac{1}{b+c}+\frac{1}{d+a}\right)}$$

WLOG $$\color{brown}{\frac{1}{b+c}+\frac{1}{d+a}} \ge \color{green}{\frac{1}{a+b}+\frac{1}{c+d}} $$

Then

$$(\color{blue}{a+c})\color{green}{\left(\frac{1}{a+b}+\frac{1}{c+d}\right)}+(\color{red}{b+d})\color{brown}{\left(\frac{1}{b+c}+\frac{1}{d+a}\right)} \ge (\color{blue}{a+c}+\color{red}{b+d})\color{green}{\left(\frac{1}{a+b}+\frac{1}{c+d}\right)}$$ $$= (\color{turquoise}{a+b}+\color{orange}{c+d})\left(\color{turquoise}{\frac{1}{a+b}}+\color{orange}{\frac{1}{c+d}}\right) \ge (1+1)^2=4$$

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    $\begingroup$ Well THAT seems very short, considering how much I actually had to type. $\endgroup$ – S.C.B. May 1 '16 at 7:22
  • $\begingroup$ Your edit was really quite clever! (+1) $\endgroup$ – Aritra Das May 1 '16 at 7:50
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    $\begingroup$ @AritraDas That's basically what it was. I don't think it can get much simpler. $\endgroup$ – S.C.B. May 1 '16 at 8:03
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    $\begingroup$ To do simpler, from line 2 use $\frac1{a+b}+\frac1{c+d}=\frac4{a+b+c+d}$ and you have it. Good one though, +1. $\endgroup$ – Macavity May 1 '16 at 11:36
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    $\begingroup$ @zxcvber It basically means without loss of generality. It is clear that one of $$\color{brown}{\frac{1}{b+c}+\frac{1}{d+a}} \ge \color{green}{\frac{1}{a+b}+\frac{1}{c+d}} $$ or $$\color{brown}{\frac{1}{b+c}+\frac{1}{d+a}} \le \color{green}{\frac{1}{a+b}+\frac{1}{c+d}} $$has to be true. However, in the second case, the proof is basically the same as in the first case. So, I without loss of generality, assumed the first one was true. $\endgroup$ – S.C.B. May 2 '16 at 8:28
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By C-S $$\sum_{cyc}\frac{a+c}{a+b}=(a+c)\left(\frac{1}{a+b}+\frac{1}{c+d}\right)+(b+d)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)\geq$$ $$\geq\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{b+c+d+a}=4.$$

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