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It's concerning this inequality from this link

The idea is to check this new inequality with $a,b,c,d,e,f$ positive real numbers where we have to find $f$ in function of the others:

$$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\geq f$$

where

$$f=\left|\frac{a-b}{b+c}\cdot\frac{b-c}{c+d}\cdot\frac{c-d}{d+e}\cdot\frac{d-e}{e+a}\cdot\frac{e-a}{a+b}\right|.$$

Thanks.

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    $\begingroup$ If you mean to find an $f$ such that the inequality holds for any $a,b,c,d,e$ then that's not possible. Suppose for example that $a \gt b$, and keep everything else constant but let $c \to b\,$, then the RHS tends to $+\infty$. $\endgroup$ – dxiv Aug 7 '17 at 23:10
  • $\begingroup$ @dxiv sorry I don't explain correctly my attempt . In fact $f$ depends entirely of $a,b,c,d,e$ .For example you could have $f=g(a,b,c,d,e)=|(a-b)(b-c)(c-d)(d-e)(e-a)|^{-1}$ .So in the case you describe $f$ can't be a constant . $\endgroup$ – user448747 Aug 8 '17 at 9:27
  • $\begingroup$ @dxiv see my edit and tell me if I'm wrong . $\endgroup$ – user448747 Aug 8 '17 at 11:06
  • $\begingroup$ Sorry, not sure what the question is after the latest edits. here I suppose that f is equal to If you prove that the inequality holds for that particular $f$ then you would have proved a stronger inequality than the one in the linked question. However, I don't see offhand any obvious reason why your inequality would hold. $\endgroup$ – dxiv Aug 8 '17 at 20:03
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I wrote a short Pascal program looking for a counterexample, and it found it with $a=2$, $b=4$, $c=6$, $d=7$, and $e=5$.

program p2385926;
const
 M=7;
var
 a,b,c,d,e:Integer;
 LHS,f:Double;
begin
for a:=1 to M do for b:=1 to M do for c:=1 to M do for d:=1 to M do for e:=1 to M do begin
 LHS:=(a-b)/(b+c)+(b-c)/(c+d)+(c-d)/(d+e)+(d-e)/(e+a)+(e-a)/(a+b);
   f:=abs((a-b)*(b-c)*(c-d)*(d-e)*(e-a)/((b+c)*(c+d)*(d+e)*(e+a)*(a+b)));
if (LHS+0.1)<f then begin writeln(a,' ',b,' ',c,' ',d,' ',e); halt end;

end;
end.
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