WakeUp wrote: Let $a,b,c,d,e,f$ be non-negative real numbers satisfying $a+b+c+d+e+f=6$. Find the maximal possible value of $\color{white}\ .\quad \ \color{black}\ \quad abc+bcd+cde+def+efa+fab$ and determine all $6$-tuples $(a,b,c,d,e,f)$ for which this maximal value is achieved. $abc+bcd+cde+def+efa+fab=(a+d)(b+e)(c+f)-ace-bdf\leq$ $\leq\left(\frac{a+b+c+d+e+f}{3}\right)^3=8$. For $a=b=c=2$ and $d=e=f=0$ it holds. For your, WakeUp. Find the maximal value of $abcd+bcde+cdef+defa+efab+fabc$, where $a,b,c,d,e,f$ be non-negative real numbers satisfying $a+b+c+d+e+f=6$.

What if $a, b, c, d, e, f > 0$?

Then the infimum is still $8$ (the minimum is not attained), as simple as that.

Arqady,can you post your solution?Thank you. I've no idea about it

DC93 wrote: Arqady,can you post your solution?Thank you. For which inequality? For the second? The answer is $6$, but my proof is very ugly and I am sure that there is something nice.