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Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that: $$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\geq0$$ Let $e=\min\{a,b,c,d,e\}$.

So, we can prove this inequality in any cases except the following two cases.

  1. $d\geq b\geq c\geq a\geq e$ and

  2. $a\geq b\geq d\geq c\geq e$.

I suggest that in this topic we'll say about these two cases only.

We have here the following things (thanx to Mida, which showed me this possibility).

Let $c\geq d$.

Thus, by C-S $$\sum_{cyc}\frac{a-b}{b+c}=\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}+$$ $$=\frac{d-c}{d+a}+\frac{a-d}{a+b}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}=$$ $$=-4+\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{a+c}{a+d}+\frac{b+d}{a+b}+$$ $$+(c-d)\left(\frac{1}{d+e}-\frac{1}{d+a}\right)+\frac{e-d}{a+b}+\frac{d-e}{e+a}=$$ $$=-4+(a+c)\left(\frac{1}{b+c}+\frac{1}{a+d}\right)+(b+d)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)+$$ $$+\frac{(c-d)(a-e)}{(d+e)(d+a)}+\frac{(d-e)(b-e)}{(e+a)(a+b)}\geq$$ $$\geq-4+\frac{4(a+c)}{b+c+a+d}+\frac{4(b+d)}{c+d+a+b}=0.$$ Id est, it's enough to prove our inequality for $d\geq c$.

Now, by using C-S for the expressions $$\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+b}+\frac{e-b}{b+c},$$ $$\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+c}+\frac{a-c}{c+d},$$ $$\frac{d-e}{e+a}+\frac{e-a}{a+b}+\frac{a-b}{b+d}+\frac{b-d}{d+e}$$ and $$\frac{e-a}{a+b}+\frac{a-b}{b+c}+\frac{b-c}{c+e}+\frac{c-e}{e+a}$$ respectively, we obtain: $(1)$ $$\sum_{cyc}\frac{a-b}{b+c}\geq\frac{(d-e)(b-a)}{(e+a)(e+b)}+\frac{(a-e)(a-c)}{(b+c)(a+b)},$$ $(2)$ $$\sum_{cyc}\frac{a-b}{b+c}\geq\frac{(a-e)(b-c)}{(a+c)(a+b)}+\frac{(a-b)(d-b)}{(b+c)(c+d)},$$ $(3)$ $$\sum_{cyc}\frac{a-b}{b+c}\geq\frac{(a-b)(d-c)}{(b+c)(b+d)}+\frac{(c-b)(c-e)}{(c+d)(d+e)}$$ and $(4)$ $$\sum_{cyc}\frac{a-b}{b+c}\geq\frac{(c-b)(d-e)}{(c+e)(c+d)}+\frac{(d-c)(d-a)}{(e+a)(d+e)}.$$ Since $d\geq c$ and $e=\min\{a,b,c,d,e\}$, it's enough to check $12$ cases and we can see that in ten of them our inequality is true.

For example, the case $a\geq d\geq b\geq c\geq e$ follows from $(2).$

Michael Rozenberg
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  • Dear @Did I think in our case it's not so duplicate because in the linked topic we have no solution. – Michael Rozenberg Aug 08 '17 at 18:20
  • @darij grinberg In the linked topic there is a very bad wrong solution. I don't want that starting topic will be duplicate. What do you think? – Michael Rozenberg Aug 08 '17 at 18:37
  • @MichaelRozenberg: Well, I still think it's a duplicate; what is gained by having two topics for it? People might well repeat the old mistaken proof here. (I assume you have looked at the literature about the Shapiro inequality for useful bits?) – darij grinberg Aug 08 '17 at 18:45
  • @darij grinberg It's something new? Or do you mean the known Shapiro's inequality? – Michael Rozenberg Aug 08 '17 at 19:07
  • I mean the known one -- I guess the literature around it might have something on this one too. – darij grinberg Aug 08 '17 at 19:15
  • @darij grinberg The convexity does not help here (the Drinfeld's stile). I tried. I tried many ways. – Michael Rozenberg Aug 08 '17 at 19:17
  • Yang Lu and Xia Bican, in their book "Automated Inequality Proving And Discovering" (World Scientific, 2016, ISBN 9814759139, 9789814759137) , have shown (see page 291, examples 10.19 and 10.20) by Successive Difference Substitution that the inequality holds, indeed it does not hold in 6 variables and it holds again in 7 variables. See here: https://books.google.de/books?id=MGC2DAAAQBAJ&pg=PA291&lpg=PA291&dq=conjecture+cirtoaje&source=bl&ots=VgwaAhwvRf&sig=dK85ICa8kt37dDTWh05vVmxyfkg&hl=de&sa=X&ved=2ahUKEwiy9q7A_5ffAhWiWhUIHamZAmw4FBDoATAAegQICRAB#v=snippet&q=cirtoaje&f=false – Andreas Dec 16 '18 at 19:14
  • @Andreas Thank you! This inequality was proven by BW and computer. In this paper there are inequalities, which we can prove without computer. For example, the inequality 10.5 has nice proof by C-S. I think using BW for the proof here it's not so good example. – Michael Rozenberg Dec 16 '18 at 21:56
  • Right. It's a confirmation that indeed a proof exists to the problem asked HERE. It doesn't say how this proof looks like, when written down. It also doesn't say whether there is a nicer proof. – Andreas Dec 16 '18 at 22:15

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