sqing wrote: Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$ Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Prove that$$-512\leq(a+b)(b+c)(c+d)(d+e)(e+a)\leq 288.$$

WAIT... SOURCE: CHENGDU NOVEMBER 14, 2018... I LIVE IN CHENGDU, AND IT IS NOVEMBER 14, 2018 I mean WHAT????

codeninja29 wrote: WAIT... SOURCE: CHENGDU NOVEMBER 14, 2018... I LIVE IN CHENGDU, AND IT IS NOVEMBER 14, 2018 I mean WHAT???? The 2018 CMO is now hosted at Chengdu No.7 High School. Today is the first day of contest, and tomorrow would have one more contest.

I had no idea...

sqing wrote: Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Prove that$$-512\leq(a+b)(b+c)(c+d)(d+e)(e+a)\leq 288.$$ Let $a_1,a_2,\cdots,a_{2k+1}\geq -1 $ and $a_1+a_2+\cdots +a_{2k+1}=2k+1$ $(k\in N^+).$ Prove or disprove $$-2^{2k+3}\cdot k^2\leq (a_1+a_2)(a_2+a_3)\cdots (a_{2k}+a_{2k+1})(a_{2k+1}+a_1)\leq \max\{2^{2k+1}\cdot (2k-1)^2,2^{2k}\cdot k(2k-1)^2\}.$$Let $a_1,a_2,\cdots,a_{2k}\geq -1 $ and $a_1+a_2+\cdots +a_{2k}=2k$ $(2\leq k\in N^+).$ Prove or disprove $$-2^{2k}\cdot (2k-1)(k-1)^2\leq (a_1+a_2)(a_2+a_3)\cdots (a_{2k-1}+a_{2k})(a_{2k}+a_1)\leq 2^{2k}\cdot (2k-1)^2.$$(Zhangyanzong)

Sketch of a terrible solution: casework on number of variables <2 after making substitution $v=a+b+2,w=b+c+2$, etc. AM-GM to increase absolute value of S, this absolute max is reached in each case when sum of numbers < 2 is equal to 0. Equality when 3,4 variables of v,w,x,y,z are =0.

sqing wrote: sqing wrote: Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$ Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Prove that$$-512\leq(a+b)(b+c)(c+d)(d+e)(e+a)\leq 288.$$ WLOG suppose that $e=\max\{a,b,c,d,e\}\Longrightarrow d+e\ge0, e+a\ge0$. To establish the minimum value (<0) of S, we have 2 possibilities: m1) Only one of $a + b$, $b + c$, $c + d$ is negative, suppose $a + b\le 0$, then by AM-GM Inequality $$-S=(-a-b)(b+c)(c+d)(d+e)(e+a)\le \left(\frac{2(c+d+e)}{5}\right)^5=\left(\frac{2(5-a-b)}{5}\right)^5\le\left(\frac{14}{5}\right)^5< 512.$$ m2) $a + b\le 0$, $b + c\le 0$, $c + d\le0$, then by AM-GM Inequality $$-S=(-a-b)(-b-c)(-c-d)(d+e)(e+a)\le \left(\frac{-a-2b-2c-d}{3}\right)^3\left(\frac{a+d+2e}{2}\right)^2=\left(\frac{a+d+2e-10}{3}\right)^3\left(\frac{a+d+2e}{2}\right)^2\le512$$for $10\le a+d+2e=10-a-2b-2c-d\le16$. Equality holds when $a=b=c=d=-1$, $e=9$. To establish the maximum value (>0) of S, we have 4 possibilities: M1) All of $a + b$, $b + c$, $c + d$ are non-negative, then by AM-GM Inequality $$S\le \left(\frac{2(a+b+c+d+e)}{5}\right)^5=32<288.$$ M2) $a+b\le0$, $b+c\le0$, $c+d\ge0$, then by AM-GM Inequality $$S=(-a-b)(-b-c)(c+d)(d+e)(e+a)\le \left(\frac{-a-2b-c}{2}\right)^2\left(\frac{c+d+e+a}{2}\right)^2(d+e)=\left(\frac{d+e-b-5}{2}\right)^2\left(\frac{5-b}{2}\right)^2(d+e)\le 288$$because $d+e+b=5-a-c\le7\Longrightarrow d+e-b-5\le2-2b$ and $(1-b)^2(5-b)^2(7-b)-4\cdot 288\le 0$ for $b\ge -1$. Equality holds when $a=b=c=-1$, $d=e=4$. M3) $a+b\ge0$, $b+c\le0$, $c+d\le0$ is similar to M2. Equality holds when $b=c=d=-1$, $a=e=4$. M4) $a+b\le0$, $b+c\ge0$, $c+d\le0$, then by AM-GM Inequality $$S=(-a-b)(b+c)(-c-d)(d+e)(e+a)\le \left(\frac{-a-b-c-d}{2}\right)^2\left(\frac{b+c+d+e+e+a}{3}\right)^3=\left(\frac{e-5}{2}\right)^2\left(\frac{5+e}{3}\right)^3< 288$$because $e\ge5$ and $b+c\ge0\Longleftrightarrow 5-a-d-e\ge0\Longrightarrow e\le 5-a-d\le7$.

MariusStanean wrote: sqing wrote: sqing wrote: Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$ Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Prove that$$-512\leq(a+b)(b+c)(c+d)(d+e)(e+a)\leq 288.$$ WLOG suppose that $e=\max\{a,b,c,d,e\}\Longrightarrow d+e\ge0, e+a\ge0$. To establish the minimum value (<0) of S, we have 2 possibilities: m1) Only one of $a + b$, $b + c$, $c + d$ is negative, suppose $a + b\le 0$, then by AM-GM Inequality $$-S=(-a-b)(b+c)(c+d)(d+e)(e+a)\le \left(\frac{2(c+d+e)}{5}\right)^5=\left(\frac{2(5-a-b)}{5}\right)^5\le\left(\frac{14}{5}\right)^5< 512.$$ m2) $a + b\le 0$, $b + c\le 0$, $c + d\le0$, then by AM-GM Inequality $$-S=(-a-b)(-b-c)(-c-d)(d+e)(e+a)\le \left(\frac{-a-2b-2c-d}{3}\right)^3\left(\frac{a+d+2e}{2}\right)^2=\left(\frac{a+d+2e-10}{3}\right)^3\left(\frac{a+d+2e}{2}\right)^2\le512$$for $10\le a+d+2e=10-a-2b-2c-d\le16$. Equality holds when $a=b=c=d=-1$, $e=9$. To establish the maximum value (>0) of S, we have 4 possibilities: M1) All of $a + b$, $b + c$, $c + d$ are non-negative, then by AM-GM Inequality $$S\le \left(\frac{2(a+b+c+d+e)}{5}\right)^5=32<288.$$ M2) $a+b\le0$, $b+c\le0$, $c+d\ge0$, then by AM-GM Inequality $$S=(-a-b)(-b-c)(c+d)(d+e)(e+a)\le \left(\frac{-a-2b-c}{2}\right)^2\left(\frac{c+d+e+a}{2}\right)^2(d+e)=\left(\frac{d+e-b-5}{2}\right)^2\left(\frac{5-b}{2}\right)^2(d+e)\le 288$$because $d+e+b=5-a-c\le7\Longrightarrow d+e-b-5\le2-2b$ and $(1-b)^2(5-b)^2(7-b)-4\cdot 288\le 0$ for $b\ge -1$. Equality holds when $a=b=c=-1$, $d=e=4$. M3) $a+b\ge0$, $b+c\le0$, $c+d\le0$ is similar to M2. Equality holds when $b=c=d=-1$, $a=e=4$. M4) $a+b\le0$, $b+c\ge0$, $c+d\le0$, then by AM-GM Inequality $$S=(-a-b)(b+c)(-c-d)(d+e)(e+a)\le \left(\frac{-a-b-c-d}{2}\right)^2\left(\frac{b+c+d+e+e+a}{3}\right)^3=\left(\frac{e-5}{2}\right)^2\left(\frac{5+e}{3}\right)^3< 288$$because $e\ge5$ and $b+c\ge0\Longleftrightarrow 5-a-d-e\ge0\Longrightarrow e\le 5-a-d\le7$. Thanks.

Long casework, but whatever. Suppose one of $(a+b), (b+c), (c+d), (d+e), (e+a)$ is $0$, then the quantity is obviously zero. We will now consider each of these not being zero. We claim that the maximum is $288$, which is achieved when $(4,4,-1,-1,-1)$. CASE 01. Suppose that all of them are positive. Let $e+a = \min\{ (a+ b), (b+c), (c+d), (d+e), (e+a) \}$. Therefore, $e + a \le 2$. Furthermore, this gives us $a, e \le 3$. \begin{align*} S &= (a+b)(b+c)(c+d)(d+e)(e+a) \\ &= (a+b)(c+d)(b+c)(d+e)(e+a) \\ &\le \left( \frac{a+b+c+d}{2} \right)^2 \left( \frac{b+c+d+e}{2} \right)^2 (e + a) \\ &= \frac{1}{16} (5 - e)^2 (5 - a)^2 (e + a) \\ &\le \frac{1}{32} \left( \frac{2(5-e) + 2(5 - a) + 2(e+a)}{5} \right)^5 \\ &= \frac{1}{32} \cdot 4^5 \\ &= 32 \end{align*} CASE 02. Suppose that three of them are positive. The other two are negative. There are two cases: Either both of the negatives lie next to each other or not. For the first case, we could WLOG $ a + b < 0, c + d < 0$. Now notice that \begin{align*} S &= (a+b)(b+c)(c+d)(d+e)(e+a) \\ &= (a+b)(c+d)(b+c)(d+e)(e+a) \\ &\le \left( \frac{a+b+c+d}{2} \right)^2 \left(\frac{b+c+d+e}{2} \right)^2 (e+a)\\ &= \left( \frac{e - 5}{2} \right)^2 (b+c)(d+e)(e+a) \\ &=\frac{1}{4} (e - 5)(e-5)(b+c)(d+e)(e + a) \\ &\le \frac{1}{4} \left( \frac{3e - 5}{5} \right)^5 \\ &< 84 \end{align*}The last inequality follows since $e < 7$ which is from $(a + d) + (b+c) + e = 5$. Notice that $a + b + c + d < 0$, and $b + c > 0$. For the second case, we could WLOG $ a + b < 0, b + c < 0$. Now notice that \begin{align*} S &= (a+b)(b+c)(c+d)(d+e)(e+a) \\ &\le \left( \frac{a+2b+c}{2} \right)^2 (c + d)(e+a)(d+e) \\ &= \left( \frac{a+2b+c}{2} \right)^2 \left( \frac{c+d+e+a}{2} \right)^2 (d + e) \\ &= \frac{1}{16} (5 + b - d - e )^2 (5 - b)^2 (d + e) \end{align*}Notice that $0 < d + e = 5 - a - b - c \le 8$. Furthermore, $-1 \le b \le 9$, and $ b + d + e = 5 - a - c \le 7$. and since \[ f(x,y) = (5 + x - y)^2 (5 - x)^2 y \]is a convex function. Therefore, we just need to check several cases: and the maximum is clearly achieved when $d + e = 8$ and $b = -1$. This gives us $a = -1$ and $c = -1$ as well, which gives us the maximum value of $288$. CASE 03. Suppose that only one of them is positive. The other four are negative. This is impossible as we must have WLOG $a + b > 0$, but summing up $(b+c) + (c+d) + (d+e) + (e+ a) < 0$, we have $ a + b > 10$, which is impossible as $a + b = 5 - c - d -e \le 8$. Now, I claim the minimum is $-512$, achieved when $(a,b,c,d,e) = (9,-1,-1,-1,-1)$. Notice that there are 3 case CASE 01. All of them are negative. This is impossible, as their sum must be negative as well. CASE 02. 2 of them are positive. There are two cases here as well: We could WLOG either both $a+b > 0$ and $b + c> 0$ or $a+b> 0$ and $c+d > 0$. We consider the case where $a+b > 0$ and $b + c > 0$ first Now, furthermore notice that this equation implies that \begin{align*} -S &= (a+b)(b+c)(-c-d)(-d-e)(-e-a) \\ &\le \left( \frac{a+2b+c}{2} \right)^2 \left( \frac{-c-2d-2e-a}{3} \right)^3 \\ &= \left( \frac{5 + b - d - e}{2} \right)^2 \left( \frac{b - d - e - 5}{3} \right)^3 \end{align*}Where we have $b - d - e \le 5 - a - c - 2d - 2e \le 11$. Let $k = b - d - e$, we have and thus, \[ S = \left( \frac{5 + k}{ 2} \right)^2 \left( \frac{k - 5}{3} \right)^3 \le 512 \]Equality achieved when $a = c = d = e = -1$ and $b = 9$. Now, for the case where $a + b > 0$ and $c + d > 0$. \begin{align*} -S &= (a+b)(c+d)(-b-c)(-d-e)(-e-a) \\ &\le \left( \frac{a+b+c+d}{2} \right)^2 (-b-c)(-d-e)(-e-a) \\ &= \frac{1}{4} (5 - e)(5 - e)(-b-c)(-d-e)(-e-a) \\ &\le \frac{1}{4} \left( \frac{5 - 3e}{5} \right)^5 \\ &< 3 \end{align*}Since $e \ge -1$. CASE 03. 4 of them are positive. We could WLOG $a+b < 0$. Now, notice that we wanted to find the maximum value of the expression \[ -S = (-a-b)(b+c)(c+d)(d+e)(e+a) \]where all of $-a-b, b+c, c+d, d+e, e+a$ are positive. Therefore, \begin{align*} -S &= (-a-b)(b+c)(c+d)(d+e)(e+a) \\ &\le \left( \frac{2(c+d+e)}{5} \right)^5 \\ &\le \left( \frac{14}{5} \right)^5 \\ &< 512. \end{align*}

sqing wrote: Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Prove that$$-512\leq(a+b)(b+c)(c+d)(d+e)(e+a)\leq 288.$$ Let $a,b,c,d,e \geq 0$ and $a+b+c+d+e=5.$ Prove that $$(a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+e^2)(e^2+a^2) \leq \frac{729}{2}$$

EDIT: Forget this, incorrect. Equality is not preserved.

sqing wrote: sqing wrote: Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Prove that$$-512\leq(a+b)(b+c)(c+d)(d+e)(e+a)\leq 288.$$ Let $a,b,c,d,e \geq 0$ and $a+b+c+d+e=5.$ Prove that $$(a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+e^2)(e^2+a^2) \leq \frac{729}{2}$$ Let $a, b,c,d,e$ be nonnegative numbers such that $a+b+c+d+e =5.$ Prove that$$(a^2-ab+b^2)(b^2-bc+c^2)(c^2-cd+d^2)(d^2-de+e^2)(e^2-ea+a^2)\leq 192.$$(GG)

sqing wrote: Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$ In this proof we will refer to $a+b, b+c, c+d, d+e, e+a$ as ``terms". We claim that the minimum and maximum are $-512$ and $288$, achieved at $(-1, -1, -1, -1, 9)$ and $(-1, -1, -1, 4, 4)$, respectively. We first note that if one of the terms is zero, the product will be zero. We henceforth assume that none of the terms are zero. We proceed with casework on the signs of the terms. Clearly, to get the minimum, we want an odd number of the terms to be negative. Assume that exactly one of the terms is negative (WLOG, $a+b$). Then \begin{align*} (a+b)(b+c)(c+d)(d+e)(e+a) & = -\left((-a-b)(b+c)(c+d)(d+e)(e+a)\right)\\ & \ge -\left(\frac{(-a-b) + (b+c) + (c+d) + (d+e) + (e+a)}{5}\right)^5\\ & = -\left(\frac{2(c+d+e)}{5}\right)^5\\ & = -\left(2 \cdot \frac{5-a-b}{5}\right)^5\\ & \ge -\left(2 \cdot \frac{7}{5}\right)^5\\ & = -\left(\frac{14}{5}\right)^5 \end{align*} This is more than $-512$. If three of the terms are negative, we have two cases. Either all three of the said terms are ``adjacent" in the product, or all two of the said terms are ``adjacent" and one isn't. In the first case, assume WLOG that $a+b, b+c,$ and $c+d$ are negative. Then \begin{align*} (a+b)(b+c)(c+d)(d+e)(e+a) & = -\left((-a-b)(-b-c)(-c-d)(d+e)(e+a)\right)\\ & \ge -\left(\left(\frac{(-a-b) + (-b-c) + (-c-d)}{3}\right)^3 \left(\frac{(d+e) + (e+a)}{2}\right)^2\right)\\ & = -\left(\left(\frac{-a-2b-2c-d}{3}\right)^3 \left(\frac{2e+a+d}{2}\right)^2\right)\\ & = -\left(\left(\frac{e-b-c-5}{3}\right)^3 \left(\frac{5+e-b-c}{2}\right)^2\right).\\ \end{align*} Then since $e-b-c = 5-a-2b-2c-d \le 11$, we have \begin{align*} -\left(\left(\frac{e-b-c-5}{3}\right)^3 \left(\frac{5+e-b-c}{2}\right)^2\right) \ge -512.\\ \end{align*} In the second case, assume WLOG that $a+b, b+c,$ and $d+e$ are negative. Then \[(a+b) + (b+c) + (d+e) = 5+b\] is also negative. However, this is a contradiction since $b \ge -1$. Finally, we clearly cannot have all five terms negative, since the sum of the terms is $10$. This proves the minimum. To prove the maximum, we apply a similar method. Clearly, an even number of the terms must be negative, so we apply casework. If none of the terms are negative, then \begin{align*} (a+b)(b+c)(c+d)(d+e)(e+a) & \le \left(\frac{(a+b) + (b+c) + (c+d) + (d+e) + (e+a)}{5}\right)^5\\ & = 32. \end{align*} If two of the terms are negative, we apply casework on whether or not the terms are ``adjacent" in the product. If the terms are adjacent, WLOG let them be $a+b$ and $b+c$. Then \begin{align*} (a+b)(b+c)(c+d)(d+e)(e+a) & = (-a-b)(-b-c)(c+d)(d+e)(e+a)\\ & \le \left(\frac{(-a-b) + (-b-c)}{2}\right)^2 \cdot \left(\frac{(c+d) + (e+a)}{2}\right)^2 \cdot (d+e)\\ & = \left(\frac{-a-2b-c}{2}\right) \cdot \left(\frac{5-b}{2}\right)^2 \cdot (5-a-b-c).\\ \end{align*} Then we see that $-a-2b-c \le 4$, $5-b \le 6$, and $5-a-b-c \le 8$, so \begin{align*} \left(\frac{-a-2b-c}{2}\right) \cdot \left(\frac{5-b}{2}\right)^2 \cdot (5-a-b-c) \le 288. \end{align*} If the terms are not adjacent in the product, WLOG let them be $a+b$ and $c+d$. Then \begin{align*} (a+b)(b+c)(c+d)(d+e)(e+a) & = (-a-b)(b+c)(-c-d)(d+e)(e+a)\\ & \le \left(\frac{(-a-b) + (-c-d)}{2}\right)^2 \cdot \left(\frac{(b+c) + (d+e) + (e+a)}{3}\right)^3\\ & = \left(\frac{e-5}{2}\right)^2 \cdot \left(\frac{e+5}{3}\right)^3.\\ \end{align*} Then since $b + c > 0,$ we have $5-a-d-e > 0,$ or $e < 5-a-d \le 7.$ Therefore, \begin{align*} \left(\frac{e-5}{2}\right)^2 \cdot \left(\frac{e+5}{3}\right)^3 & < 256 \end{align*} which is less than $288$. Finally, if four of the terms are negative, WLOG let them be $a+b, b+c, c+d, d+e$. Then \[(a+b) + (c+d) + (d+e) = 5+d\] is also negative. This is clearly a contradiction, since $d \ge -1$. This completes the upper bound, and therefore, the problem. $\square$

The minimum and maximum are -512 and 288 achieved by $(-1, -1, -1, -1, 9)$ and $(-1, -1, -1, 4, 4)$. Now we will prove these bounds are best. The rest of the solution is a bunch of casework on the signs of $a+b, b+c, \ldots, e+a$. Let a string $\pm\pm\pm\pm\pm$ denote these signs, in order. If the product is $+++++$, then by AM-GM the max is 32. If the product is $-++++$, then smoothing $(a, e)$ and $(b, c)$ gives $a=b=-1$. By AM-GM, the maximum of $(c-1)(c+d)(d+e)(e-1)$ is $81$ and $-162 > -512$. If the product is $--+++$, smoothing $(a, e)$, $(b, d)$, and $(c, d)$ gives $a=b=c=-1$. By AM-GM and $d+e=8$, we have $4(d-1)(d+e)(e-1) = 32(d-1)(e-1) \le 32\cdot3^3 = 288$. If the product is $-+-++$, we can smooth $(a, e)$ and $(d, e)$ to get $a=d=-1$. But $b, c < 1$, so we can establish an upper bound of $(-2)2(-2)(e-1)(e-1)$. Since $b+c > 0$, we have $e < 7$, so this is less than $8\cdot 36 = 288$. If the product is $---++$, smoothing gives $(-1, -1, -1, -1, 9)$. $-+--+$ is impossible because then $a, b, c, d, e < 1$. It's clear that $----+$ and $-----$ are impossible, so we're done.

Let $a,b,c,d,e \geq 0$ and $a^2+b^2+c^2+d^2+e^2=16.$ Prove that$$a+b+c+d+e \leq 8+\frac{1}{16}abcde$$

Least contrived Chinese inequality... The minimum and maximum are $-512,288$ achieved at $(9,-1,-1,-1,-1),(4,-1,-1,-1,4)$ respectively. First, I will show that at least two of $a+b,b+c,c+d,d+e,e+a$ are positive. WLOG let $\max\{a,b,c,d,e\}=a$. Clearly if one of $a,b,c,d,e$ is negative, then $a>1$, in which case $a+b,e+a>0$ as desired. Otherwise, it is evident that $a+b,e+a$ are positive, as desired. I will now show that $-512 \leq (a+b)(b+c)(c+d)(d+e)(e+a)$. Suppose the product $(a+b)(b+c)(c+d)(d+e)(e+a)$ is negative. Then either one or three of the terms are negative, by the above claim. We consider these two cases separately. Case 1: One of the terms is negative. Then, WLOG let $a+b$ be negative. By AM-GM, it follows that $$(-a-b)(b+c)(c+d)(d+e)(e+a)\leq \left(\frac{2(c+d+e)}{5}\right)^5\leq \left(\frac{14}{5}\right)^5$$Since $(\tfrac{14}{5})^5<3^5=243<512$, it follows that $(a+b)(b+c)(c+d)(d+e)(e+a)>-512$ in this case. Case 2: Three of the terms are negative. WLOG let $\max\{a,b,c,d,e\}=a$, so $a+b,e+a$ are positive and the rest are negative. By AM-GM, it follows that \begin{align*} (a+b)(-4b-4c)(-4c-4d)(-4d-4e)(e+a)&\leq \left(\frac{2a-3b-3e-8c-8d}{5}\right)^5\\ &\leq \left(\frac{18+3+3+8+8}{5}\right)^5=8^5=2^{15}, \end{align*}which implies $(a+b)(b+c)(c+d)(d+e)(e+a)\geq -2^{15}/2^6=-512$, as desired. Now, I will show that $(a+b)(b+c)(c+d)(d+e)(e+a) \leq 288$. Suppose $(a+b)(b+c)(c+d)(d+e)(e+a)$ is positive. Then with our claim, we can consider the following two cases. Case 1: None of the terms are negative. Then from AM-GM we have $$(a+b)(b+c)(c+d)(d+e)(e+a)\leq \left(\frac{2(a+b+c+d+e)}{5}\right)^5=32,$$which implies $(a+b)(b+c)(c+d)(d+e)(e+a)\leq 288$. Case 2: Two of the terms are negative. If the two terms share a variable (for example, $a+b$ and $e+a$), then WLOG let this shared variable be $a$. We then have (by AM-GM) \begin{align*} (-12a-12b)(8b+8c)(3c+3d)(8d+8e)(-12e-12a)&\leq \left(\frac{-24a-4b-4e+11c+11d}{5}\right)^5\\ &\leq \left(\frac{24+4+4+11(8)}{5}\right)^5=24^5, \end{align*}which gives $(a+b)(b+c)(c+d)(d+e)(e+a)\leq 288$, the desired bound. If the two terms don't share a variable, then WLOG let them be $a+b$ and $c+d$. In this case, we have \begin{align*} (-2a-2b)(2b+2c)(-2c-2d)(d+e)(e+a)&\leq \left(\frac{-a-d+2e}{5}\right)^5\\ &\leq \left(\frac{1+1+18}{5}\right)^5=4^5 \end{align*}hence $(a+b)(b+c)(c+d)(d+e)(e+a)\leq 1024/8=128\leq 288$, as desired. $\blacksquare$

I'll prove that \[ -512 \leq (a+b)(b+c)(c+d)(d+e)(e+a). \leq 288 \] The lower bound is achieved for example when $(a,b,c,d,e)=(-1;-1;-1;-1;9);(-1;-1;5;-1;5)$ and the upper bound when $(a,b,c,d,e)=(-1;-1;-1;4;4)$. First off, if WLOG $b$ is the largest, then $a+b$ and $b+c$ are positive, so there are at most 3 negative terms in the product. Since negative values are achievable, the minimum of the product is achieved either when one or three of them are negative. Assume they are three, the other case is analogous. If the sum of the negative values is $-x$, then since the sum of all of the terms is $10$, the sum of the positive values is $10+x$. By AM-GM once for negative and once for the positive terms we get that the mininal value is $-(\frac{x}{3})^3(\frac{10+x}{2})^2$. Since all of the letters are $\geq -1$, then all of the terms are $\geq -2$ , thus $x\leq 6$ and the minimum is achieved when $x=6$ and it's $-512$. If only one is negative, then the minimum is $-x3(\frac{10+x}{4})^4$, where $x\leq 2$ and the min is again $-512$ somehow. My way of providing the bound for the maximum is quite bashy. If no of the terms are negative, then AM-GM straight up gives an upper bound of $32$. Thus exactly two of them should be negative. Now some casework for which two should be negative and then smoothing, in which we make certain letters $-1$ and other larger, while increasing the product, does the job.

We claim that: \[-512\leq S\leq 288\]The equality cases are $(9,-1,-1,-1,-1)\to 8^2\cdot (-2)^3 = -512$ and $(4,4,-1,-1,-1)\to 8\cdot 3^2\cdot (-2)^2 = 288$. To prove that these bounds are valid, we conduct a casework analysis. Case 1: 5 of the values from $\{a+b,b+c,c+d,d+e,e+a\}$ are negative. This case is clearly impossible since the sum of the elements of the set is 10. $\square$ Case 2: 4 of the values from $\{a+b,b+c,c+d,d+e,e+a\}$ are negative. Assume that $a+b$ is positive. Then, since the sum if 10, we must have that $a+b\geq 10$. Then, WLOG $a\geq 5$. This implies that $e+a\geq 5-1>0$ must also be positive, so this case is also impossible. $\square$ Case 3: 3 of the values from $\{a+b,b+c,c+d,d+e,e+a\}$ are negative. In this case, note that the final product $x_1x_2x_3x_4x_5$ must be negative, so it suffices to bound the magnitude of this. To do this, note that if $x_i<0$ and $x_j>0$, we may ``smooth", decreasing $x_i$ to -1 and increasing $x_j$ by a positive amount, which will increase both magnitudes while preserving all conditions. Thus, the final optimal configuration will have $x_1=x_2=x_3=-2$ and $x_4+x+5=16$. This is clearly maximized when $x_4=x_5=8$, yielding $(-2)^3\cdot 8^2 = -512$. $\square$ Case 4: } 2 of the values from $\{a+b,b+c,c+d,d+e,e+a\}$ are negative. There are two cases. Also, note that the value is always positive. In the first subcase, the two negative sums are disjoint; WLOG $a+b<0,c+d<0$, and all else are positive. This implies that $|a+b|+|b+c|+|c+d| = -(a+b) + (b+c) - (c+d) = -a-d\leq 2$. Thus, $|(a+b)(b+c)(c+d)|\leq \left(\frac23\right)^3$. Then, $(d+e)+(e+a) \leq 10 + 4 = 14$, so $|(d+e)(e+a)|\leq 7^2$. Putting it together, the product is at most \[\frac{8}{27}\cdot 49 < 8\cdot 2 < 288\] In the second subcase, we have overlapping negative sums; WLOG $a+b<0, b+c<0$. Then, we may smooth as before since any of the following mappings will increase $S$: \begin{align*} (a,b,c,d,e)&\to (a-\epsilon, b,c,d,e+\epsilon)\\ (a,b,c,d,e)&\to (a, b-\epsilon,c,d,e+\epsilon)\\ (a,b,c,d,e)&\to (a, b,c-\epsilon,d+\epsilon,e)\\ \end{align*}until we have $a=b=c=-1$. At this point, $d+e=8$, and the maximum value is \[(a+b)(b+c)(c+d)(d+e)(e+a) = -2^2\cdot 8\cdot (d-1)(e-1) \leq 32\cdot 3^2 = 288\]$\square$ Case 5: 1 of the values from $\{a+b,b+c,c+d,d+e,e+a\}$ are negative. Using similar logic as Case 3, we can smooth to $x_1=-2$ and $x_2+x_3+x_4+x_5=12$, which has maximum magnitude $2\cdot 3^4 = 162 < 512$. $\square$ Case 6: none of the values from $\{a+b,b+c,c+d,d+e,e+a\}$ are negative. In this case, all are positive, and all five values add to 10, so the maximum is $2^5 =32< 288$.$\square$ Thus, we have shown that $-512\leq S\leq 288$, with achievable bounds, and have therefore successfully found the minimum and maximum values. $\blacksquare$.

We claim the minimum possible value of $S$ is $-512$ and the maximum possible value of $S$ is $288$. These are attainable by $(-1,-,1,-1,-1,9)$ and $(-1,-1,-1,4,4)$, respectively. First we prove that $S \ge -512$. Consider some $a,b,c,d,e$ achieving the minimum value of $S$, and suppose that the minimum was less than $-512$. Clearly an odd number of $a+b, b+c, c+d, d+e, e+a$ is negative. Since we can't have all five negative (they add up to $10$), one or three of them are negative. Case 1: One of them is negative. WLOG $a + b$ is negative. Notice that $(b+c) +(c+d) + (d+e) + (e+a) \le 12$, and $(b+c), (c+d), (d+e), (e+a)$ are all positive, so $(b+c)(c+d)(d+e)(e+a) \le 3^4 = 81$ by AM-GM. Now, we have $(a+b) \ge -2$, meaning that $|S| \le 81\cdot 2 = 162$, so $S \ge -162$ in this case. Case 2: Three of them are negative. Therefore two of these terms must be positive. Since all five add up to $10$, the product of sum two positive terms is at most $16$, so their product is at most $64$ by AM-GM. The product of the other three terms is a negative number at least $-8$. Hence $|S| \le 64 \cdot 8 = 512$, meaning that $S \ge -512$. Now we show that $S\le 288$. Suppose some value of $S$ was over $288$. Clearly an even number of $a+b, b+c, c+d, d+e, e+a$ is negative. Case 1: None of them are negative. Since they add up to $10$ and are positive, their product is at most $2^5 = 32 < 288$. Case 2: Two of them are negative. Subcase 2.1: If the two negative terms didn't have any of $a,b,c,d,e$ inside both of them then WLOG that they are $a+b$ and $c + d$. Let $b + c = 2x$ for some positive real number $x$. We have $(b + c) + (d + e) + (e + a) \le 14$, so $(d + e) + (e + a) \le 14 - 2x$. This implies that $(d + e) (e + a) \le (7-x)^2$ by AM-GM. Now, since $a + b$ and $c + d$ are negative, but $b + c$ is positive, it means that all of $a,b,c,d$ are less than $1$, so $x < 1$. Now we have\[S = (a+b)(c+d) \cdot 2x (7-x)^2 \le 8x \cdot (7-x)^2,\]which is less than $288$ for $0 < x < 1$. Subcase 2.2: Otherwise, WLOG $a + b$ and $b + c$ were the two negative terms. Let $a' = -1, b' = -1, c' = -1, d' = c+ d + 1, e' = a + e + 1$ and\[ S' = (a' + b') (b' + c') (c' + d') (d' + e') (e' + a') \]Notice that $a' + b', b' + c' $ are negative and $c' + d', d' + e' , e' + a' $ are positive. We have $|a' + b'| > |a+b|, |b' + c'| > |b+c| , |c' + d'| = |c + d|, |d' + e'| > |d + e|, |e' + a'| = |e + a|$, hence\[ |S'| > |a+b| \cdot |b+c| \cdot |c+d| \cdot |d+e| \cdot |e + a| = |S| > 288\]Since $S' $ is positive, we must have $S' > 288$. However, note that $d' + e' = a + c + d + e + 2 = 7-b \le 8$. Now we see that if $d'$ and $e'$ were greater than $1$, we have\[(d' - 1) (e' - 1) \le \left( \frac{d' +e' - 2}{2} \right)^2 \le 9 \]Otherwise, $(d' - 1) (e' - 1) < 9$ obviously. Hence\[ S' = 4 (d' - 1)(e' - 1) (d' + e') \le 4\cdot 9 \cdot 8 = 288,\]contradiction. Case 3: Four of them are negative. The four negative terms have a product of at most $16$ and the one positive term is at most $10 + 2\cdot 4 = 18$, so the product of the five terms is at most $16\cdot 18 = 288$.

first for the lower bound, we show it is $-512$ when $a=b=c=d=-1,e=9.$ this is by casework: when there are $1$ negatives $a+b,b+c,c+d,d+e,e+a$ it is $\ge -2$ so the others sum to $12$ and by amgm product minimized at $-2\cdot 3\cdot 3\cdot 3\cdot 3=-162,$ when there are $3$ negatives they are $\ge -2$ and the other $2$ sum to $16$ so amgm gives $-2\cdot -2\cdot -2\cdot 8\cdot 8=-512$ the bound, now $5$ negatives is impossible so this case is done. for the upper bound we claim $288$ when $a=b=c=-1,d=e=4.$ if there are $0$ negatives in $a+b,b+c,c+d,d+e,e+a$ amgm gives $\le 32.$ if there are $4$ negatives they are $\ge -2$ and the last one is $\le 18$ so their product is $\le-2\cdot -2\cdot -2\cdot -2\cdot 18=288.$ now if there are $2$ negatives first suppose they are not adjacent, that is suppose $a+b,c+d$ are negative and the others positive. if we decrease $a$ by some amount and increase $e$ by the same amount the product increases, so wlog $a=-1,$ similarly $d=-1.$ now $|b+c|+|a+b|+|c+d|=-(a+d)=2$ so amgm gives $(a+b)(b+c)(c+d)\le\tfrac8{27}.$ then $d+e,e+a$ sum to $\le 14$ so their product is $\le 49$ and overall its $<288.$ finally if $a+b,b+c$ are negative and the others positive, decreasing $a$ and increasing $e$ by the same amount lets us wlog $a=-1,$ same with $c=-1,$ then decreasing $b$ and increasing $e$ lets us $b=-1.$ now we have $d+e=8,$ amgm says $(d-1)(e-1)\le9$ so together we have $(a+b)(b+c)(c+d)(d+e)(e+a)\le-2\cdot-2\cdot8\cdot9=288$ done.

I throw around the term absolute value bounding, which basically means take each factor and replace it with its optimal absolute value, or something higher for tractability. Also I use $a,b,c,d,e \le 9$ without mention. Also whenever I say "take" something to something else the sum of $a + b + c + d +e = 5$ is always preserved. I claim that the maximum is $288$, achieved by $(a,b,c,d,e) = (-1,-1,-1,4,4)$. We do casework on the positivity of $a,b,c,d,e$, with zero being counted as positive for laziness purposes. Case 1: All are negative. This is clearly impossible as the sum is positive. Case 2: $4$ are negative, $1$ is positive. Let $a$ be positive, then we see that since $c + d + e$ is negative, we have that $a + b$ is positive, and so is $a + e$, and $b + c$, $c + d$, $d + e$ are all negative, so the product is negative overall. Less than $288$. Case 3: $3$ are negative, $2$ is positive. We break this into two subcases, one where the three negative elements are all contiguous and one where they aren't. In the first case, let $a,b,c$ be negative, and $d,e$ be positive. We strictly increase the absolute value and maintain the sign by sending $c$ to $-1$ and $d$ to $c + d + 1$, as $d + e$ increases and it is positive and $b + c$ decreases and it is negative. Likewise, we send $a$ to $-1$. Now we want to optimize $(b- 1)^2 (d - 1)(d + e)(e - 1)$, then observe $b + d + e = 7$, so we can conveniently write $b = 7 - d- e$ and get $(6 - d - e)^2(d + e)(d - 1)(e - 1)$. We know that $d,e \ge 1$. Now let $d + e = x$, then we wish to optimize $(6 - x)^2 (x)(d -1)(e - 1)$, optimizing $(d - 1)(e -1)$ is optimizing the product of two nonnegative numbers summing to $x - 2$, so this is just $\frac 14(x -2)^2$. The entire expression we wish to optimize is then $\frac 14 (6 - x)^2 x (x-2)^2$, where $7 \le x \le 8$. Now this is clearly increasing over that interval, so we just take $x = 8$ and get the answer $288$. In the second case, let $a,c,e$ be negative and $b,d$ be positive. Since $(a + e)$ is negative, another one of the sums also must be negative. Let it be one of $a + b$, $b + c$, this then forces $b \le 1$, taking absolute values the maximum is $2d^2$, clearly less than $288$. Case 4: $3$ are positive, $2$ is negative. We break this into two subcases, one where the three positive elements are all contiguous and one where they aren't. In the first case, take $a,b,c$ positive and $d,e$ negative. Change $d$ to $-1$ and $c$ to $c + d + 1$. Now $b + c$ strictly increases and remains positive, $c + d$ remains the same, $d + e$ decreases but it is already negative, so the sign of the product remains the same and the absolute value increases. We are now being asked to optimize $(a + b)(b + c)(c -1)(-2)(a - 1)$. We use an absolute value argument (all arguments for this section conveniently hold no matter what the sign of the product is). Write $a + b + c = 7$, so substituting gives $(-2)(7 - c)(c -1)(7 - a)(a - 1)$. We can optimize each quadratic in $c$ and $a$ separately, to get an absolute value bound of $2 (7 - 4)^2(4 - 1)^2 = 162$, at $a = c = 4$ (note that this doesn't actually hold, since we can't use $a + c > 7$, however any working $a,c$ are still bounded by this number). This is less than $288$. In the second case, take $a,c,e$ positive, and $b,d$ negative. Clearly, if some set $a,b,c,d,e$ is maximal, we must have an even amount of $a + b$ and cyclic variants negative. We now do casework on which ones are negative. If none are negative, we can set $b = d = 0$ and retain $a,c,e$, to get $a + c + e \le 7$, maximize $(a + e)ac^2e$, which is optimized when $a + c+ e = 7$, so rewrite as $(a + e)(ae)(7 - a - e)^2$, if we fix $a + e$ the optimal value of $ae$ is $\frac{1}{4}(a+ e)^2$, so letting $a + e = x$, we want to optimize $\frac 14 x^3(7 -x)^2$. By AM-GM, we can optimize $(\frac{2x + 2x + 2x + 21 - 3x + 21 - 3x}{5})^5 \ge 72(x^3)(7 - x)^2$, so our quantity is bounded by $(\frac{42}{5})^5 \frac{1}{288}$, which is in fact less than $288$. If four are negative, this is clearly impossible since it implies $|b| > |a|, |c|$ and $|d| > |e|$, summing forces the total sum as less than $|b|$, so less than $1$. Thus we turn to the case where there are two negatives. If $b + c$, $c + d$ are the negatives, send $b$ to $0$ and $c$ to $b + c$, so the total product changes by a factor of $\frac{b + c + d}{c + d} \frac{a}{a + b}$, which is greater than $1$ since $b, c+d$ are both negative, then we can treat zero as nonnegative and pretend it is the case where $4$ are positive since the arguments made always hold in the case of nonnegatives. If $a + b, b + c$ (or the symmetrical variant for $d$) are the negatives, then we can send $d$ to $0$ and $c$ to $c + d$, this changes the product by a factor of $\frac{e}{d + e} \frac{b + c + d}{b + c}$, which is greater than $1$, since both $d$, $b + c$ are greater than $1$. If $a + b$, $d + e$ are both negative, then observe $a,e$ are most $1$, and $a + b$, $d + e$ are at least $-1$, so we can bound the sum using absolute values as $2(b + c)(c + d)$, further bounded by $2c^2$, obviously less than $288$. For the last and final case, we have when $a + b$ and $c + d$ (or $b + c$ and $d + e$) are the negatives, consider $a, c, a + b, b + c, c + d$ all have magnitude less than $1$, so we are bounded by $(e + a)(e + d)$, or $(e + 1)^2$, clearly less than $288$. Ok after doing Case 3 there's probably a much faster way for that but I'm not looking. Case 5: $4$ are positive, $1$ is negative. Let $a$ be negative. We place an upper bound on the absolute value, which bounds the maximum. Clearly the absolute value of every term is increased by replacing $a$ with $-a$. Now the sum of all $5$ terms in the product is at most $14$, and since all terms are now positive, we can just use AM-GM to get a bound of $(\frac{14}{5})^5 \le 3^5 < 288$. Case 6: All are positive. This dies to AM-GM, consider $2 = \frac{\sum_{cyc} a + b}{5} $, thus $2^5$ is the upper bound of the product expression. Less than $288$. Now, we claim the minimum is $-512$, achieved by $-1, -1, -1, -1, 9$. We do casework again. Case 1: All $5$ are negative. Impossible. Case 2: $4$ are negative, $1$ is positive. As described before, this case forces a negative. Let $c$ be positive, and everything else be negative. Clearly, we benefit from sending $b$ to $-1$ and $c$ to $b + c + 1$, since the term $c + d$ goes to $b + c + d + 1$ which is greater and still positive, whereas $a + b$ goes from $a + b$ to $a - 1$, which preserves sign and causes an increase in absolute value. Likewise, changing $a$ to $-1$ and $c$ to $a + c + 1$ increases all the positive terms and decreases or keeps constant all the negative terms. Thus we can then force $a = b = d = e = -1$, giving the equality case of $-512$. Case 3: $3$ are negative and $2$ are positive. We divide this into two cases, one where the three negatives are contiguous and one where they aren't. In the first case, take $a,b,c$ negative and $d,e$ positive. We just bound absolute value without caring about signs, we see $a + b, b +c$ have optimal absolute value $2$, then $c + d$, $e + a$ have optimal absolute values $d,e$ respectively, and $d + e$ has absolute value $d + e$. Since $d + e \le 8$, we see that $de$ is optimized at $d = e = 4, de = 16$, so our total absolute value is $2 \cdot 2 \cdot 16 \cdot 8 = 512$, as desired. In the second case, take $a,c,e,$ negative and $d,e$ positive. We just bound absolute value without caring about signs. We see $a + e$ has optimal absolute value $2$, then $a + b, b + c$ has optimal absolute values of $b$, and symmetrical equivalent for $d$, so we get a maximum absolute value of $2b^2d^2$, and since $b + d \le 8$, we see $bd$ is maximized at $16$, giving the optimal absolute value at $512$, as desired. Case 4: $3$ are positive and $2$ are negative. We divide this into $3$ cases, one where the $3$ positives are contiguous and one where they aren't. In the first case, let $a,b,c$ be positive and $d,e$ be negative. Notice that taking $d$ to $-1$ and $c$ to $c + d+ 1$ increases the absolute value of the product and doesn't change the sign, symmetrically for $e,a$, so we just optimize $(a + b)(b + c)(-2)(a - 1)(c -1)$. Since $a + b + c = 7$, we can rewrite this as $(7 -a)(a - 1)(7 - c)(c - 1) (-2)$, under the constraints that $a,c$ are nonnegative. Let the function $f(x) = (7 - x)(x - 1)$, then observe $|f(4)| > |f(7)|, |f(0)|$, and $4$ is the vertex of $f$, hence the absolute value of the expression is optimized when $a = c = 4$, giving a result of $162$ for the absolute value, less than $512$. In the second case, let $a,c,e$ be positive and $b,d$ be negative. For any minimum, an odd number of sums of the form $a+b$ must be negative. Clearly all $5$ cannot be negative since $a+e$ is positive. If $3$ are negative, we see that $2$ of $a,c,e$ are at most $1$. If it is $a,c$ or $c,e$, we can bound by taking best case absolute values, we see most everything has absolute value bounded by $1$, we get $e(e + 1)$, clearly less than $512$. If it is $a,e$, we can do the same thing, except note $a + e$ has absolute value bounded by $2$ instead, so we get $2c^2$, also clearly less than $512$. Now we assume that exactly one of the sums is negative. If it is $a +b$ (or symmetrically, $d + e$), send $b$ to $-1$ and $c$ to $b + c + 1$, this strictly increases the absolute value of all terms involved. Thus we want to optimize $(a - 1)(c -1)(c + d)(d + e)(e + a)$. Note that $a$ is bounded by $1$ , $d$ bounded by $-1$, we use absolute value bounding again to an upper bound of $c^2 (d +e)(e + 1)$, at best $c + d +e =6$, we can get another upper bound as $c^2(6 - c)(e + 1)$. Now we can use AM-GM to bound $c^2(6 - c)$ as $(\frac{c + c + 12 - 2c}{3})^3 \frac 12 = 32$, then $e + 1 \le 7$, so we get a bound for absolute value of $252$. If the one negative term is $c + b$ (or symmetrically, $d + c$), send $b$ to $-1$ and $a$ to $a + b + 1$, this increases the absolute value of all terms involved and preserves signs. Thus we wish to optimize $(a - 1)(c - 1)(c + d)(d + e)(e + a)$. Absolute value bounding again, observe $c$ is at most $1$, so we can get the absolute value bound as $(a - 1)(d + 1)(d + e)(e + a)$. Finally, by AM-GM we have this is at most $(\frac{2a + 2d + 2e}{4})^4$, observe $a + c + d + e = 6$, so just take $a + d + e = 6$ at worst case, so we get $3^4$ as the bound which is less than $512$. Case 5: $4$ are positive, $1$ is negative. Let $c$ be negative. Any minimum must be negative, so we see that one of $b + c$ or $c + d$ is negative, since no other terms can be negative. Let $b + c$ be negative, then send $c$ to $-1$ and $d$ to $c + d + 1$, then we see $b + c$ decreases and it is already negative, then $d + e$ increases and it is positive, so the absolute value increases and the sign is still negative. Thus we are trying to optimize $(a + b)(b - 1)(d - 1)(d + e)(e + a)$. By absolute value optimization on each term, it suffices to optimize $(a + 1)(d - 1)(d + e)(a + e)$. By AM-GM, this is at most $(\frac{2a + 2d + 2e}{4})^4$, or at most $81$, less than $512$. Case 6: All $5$ are positive. The product is positive, so we can discard this case.

The minimum is $-512$, attained at $(-1, -1, -1, -1, 9)$. The maximum is $288$, attained at $(-1, -1, -1, 4, 4)$. We use smoothing: Claim: If $S$ is minimized or maximized, we must have $b+c = d+e$ or $-1 \in \{c, d \}$. Proof: Suppose otherwise. We claim that by varying $c$ and $d$ while fixing $c+d$, we can increase the magnitude of $S$. Indeed, as we vary $c$ and $d$, the only terms that change are $b+c$ and $d+e$. The product $(b+c)(d+e)$ varies as a quadratic function in $c$, so the product's maxima and minima are attained when $b+c = d+e$ (if applicable) and when $-1 \in \{c, d\}$, respectively. So, if $b+c \neq d+e$ and $-1 \not \in \{c, d \}$, we can smooth $c$ and $d$ to make this happen. We are left to complete a finite case check (with a hefty dose of derivatives) based off the position of the $-1$'s. This will give us the final answer.