Let 5 positive real variables $(a,b,c,d,e)$. These variables obey the conditions $a-b+c-d+e > 0$ and all 5 cyclic shifts thereof. This question is tighter than this one.
Prove: $$ \sum_{cyc} a^2 b d (c+e)\ge \sum_{cyc} a b c e (a+d) $$ where $\sum_{cyc}$ means all 5 cyclic shifts $(a,b,c,d,e) \to (b,c,d,e,a) \to$ etc. Equality occurs if all 5 variables are equal, and it appears that equality occurs at no other points. I couldn't find counterexamples through simulations.
The hint.
Let $a=x+y$, $b=y+z$, $c=z+t$, $d=t+w$ and $e=w+x$.
Thus, the condition gives that $x$, $y$, $z$, $t$ and $w$ are positives and we need to prove that $$\sum_{cyc}(x+y)^2(y+z)(t^2+w^2+zt+tw+wx-zx)\geq5\prod_{cyc}(x+y),$$ which we can prove by BW:
Let $x=\min\{x,y,z,t,w\}$, $y=x+p$, $z=x+q$, $t=x+r$ and $w=x+s.$
Practically it's better to use the last substitution for the starting inequality: $$\sum_{cyc}a^2b(cd+de-ce)\geq5abcde.$$ Id est, let $a=x+u,$ $b=x+u+v$, $c=x+v+w$, $d=x+w+p$ and $e=x+p$,
where $x>0$ and $u$, $v$, $w$ and $p$ are non-negatives.