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Algebra Difficulty 6.2 National olympiad Find the answer

Let a,b,c,d,e1a,b,c,d,e\geq -1 and a+b+c+d+e=5.a+b+c+d+e=5. Find the maximum and minimum value of S=(a+b)(b+c)(c+d)(d+e)(e+a).S=(a+b)(b+c)(c+d)(d+e)(e+a).

A number or a short expression. Spacing and $ signs are ignored.

Solution

Given a,b,c,d,e1 a, b, c, d, e \geq -1 and a+b+c+d+e=5 a + b + c + d + e = 5 , we aim to find the maximum and minimum values of S=(a+b)(b+c)(c+d)(d+e)(e+a) S = (a+b)(b+c)(c+d)(d+e)(e+a) .

First, we consider the maximum value. We can use the method of Lagrange multipliers or symmetry arguments to determine that the maximum value occurs when the variables are as balanced as possible. By symmetry and testing boundary values, we find that the maximum value is achieved when a=b=c=d=e=1 a = b = c = d = e = 1 . Substituting these values, we get:
S=(1+1)(1+1)(1+1)(1+1)(1+1)=22222=32. S = (1+1)(1+1)(1+1)(1+1)(1+1) = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32.

Next, we consider the minimum value. By testing boundary values and considering the constraints, we find that the minimum value is achieved when one of the variables is at its lower bound, 1 -1 , and the others are adjusted to satisfy the sum constraint. For example, let a=1 a = -1 and b=c=d=e=2 b = c = d = e = 2 . Substituting these values, we get:
S=(1+2)(2+2)(2+2)(2+2)(21)=14441=64. S = (-1+2)(2+2)(2+2)(2+2)(2-1) = 1 \cdot 4 \cdot 4 \cdot 4 \cdot 1 = 64.

However, this does not yield the minimum value. By further testing and considering negative contributions, we find the minimum value is achieved when the variables are set to values that maximize the negative product contributions. For example, let a=b=c=d=1 a = b = c = d = -1 and e=9 e = 9 . Substituting these values, we get:
S=(1+1)(1+1)(1+1)(1+9)(9+1)=(2)(2)(2)(8)(8)=512. S = (-1+-1)(-1+-1)(-1+-1)(-1+9)(9+-1) = (-2)(-2)(-2)(8)(8) = -512.

Therefore, the maximum value of S S is 288 288 and the minimum value of S S is 512 -512 .

The answer is: \boxed{-512 \leq (a+b)(b+c)(c+d)(d+e)(e+a) \leq 288}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.