maximum and minimum value of $(a+b)(b+c)(c+d)(d+e)(e+a).$ 
Let $a,b,c,d,e\geq  -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$

I couldn't proceed much, however I think I got the minimum and maximum case.
For minimum, we get $-512$ with equality on $(-1,-1,-1,-1,9).$
For maximum, we get $288$ with equality on   $(-1,-1,-1,4,4).$

For the maximum, you can easily prove that $4$ of the $5$ product terms cannot be negative. Then, the maximum is achieved either all $5$ are positive or only $3$ are positive. In the former case, simple $AM-GM$ gives the maximum of $32.$ For the latter case, assume $a+b\leq 0$ and $b+c\leq 0$ while the other three are non-negative. In that case, $-a-b\leq 2$ and $-b-c\leq 2$ and $d+e = 5-a-b-c\leq 8.$
Then, $$S = (c+d)(d+e)(e+a)(-a-b)(-b-c)\leq 4(c+d)(e+a)(d+e)\leq(c+d+e+a)^2\cdot 8= 8(5-b)^2\leq 8\cdot 36 = 288.$$
Now the other possibility is when two terms involving $4$ different variables are negative - say $a+b\leq 0$ and $c+d\leq 0.$ In this case, $b+c\geq 0$ and so $a+d = (a+b+c+d) - (b+c)\leq 0.$ Finally, $e = 5-a-d -(b+c)\leq 5-a-d\leq 7.$ Now for the kill:
$$S = (-a-b)(-c-d)(d+e)(e+a)(b+c)\leq \dfrac{(e-5)^2}{4}\cdot \dfrac{(e+5)^3}{27}\leq \dfrac{12^3}{27} = 64.$$
You can do similar thing for the minimum.
