I was solving the following inequality For $a$, $b$, $c$ and $d$ being positive real numbers which goes as $$ \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} \leq \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b} $$ Which I was successful able to do But I am not able to understand the intuition behind the inequality, and how someone even came to it. Can someone help me to intuitively understand it

  • One way to think about it is that, while a/(a+b) etc. are all individually less than 1, there are no similar control for the likes of a/(b+c) in the RHS (if a is relatively large compared to b and c, which just for arbitrary numbers could easily happen). Then it is just a matter of looking how far you can push it from "infinity" back and see you don't need anything more added to the RHS if you have cyclic symmetry. – user10354138 Sep 14 at 11:26
up vote 3 down vote accepted

By Cauchy-Schwarz inequality,

$$(\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}+\frac{d-a}{a+b})(b+c+c+d+a+d+a+b)\ge(a-b+b-c+c-d+d-a)^2=0$$

We need to prove the first bracket is positive, but since $a,b,c,d$ are positive, it's indeed positive.

Equality occurs iff $a-b=b-c=c-d=d-a$, i.e. $a=b=c=d.$

The intuition behind this is that $\sum_{cyc}\frac a{a+b}$ is relatively closer to $1$ than $\sum_{cyc}\frac d{a+b}$. Consider this: the sum of the numerator and the denominator are the same, but RHS is more spread out. Then which side's is bigger? This is not mathematically rigorous, but that's my first instinct. Also, these inequalities can be made by permutating Cauchy-Schwarz, rearrangement inequality, and the Power Mean Inequality.

  • I must say that Cauchy-Schwarz is so powerful... – abc... Sep 14 at 12:55

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