To prove that 20 is the smallest possible value for 4(a^2 + b^2 + c^2 + d^2) - (a+b+c+d)^2, we can use the fact that the sum of squares of consecutive integers can be expressed as (n)(n+1)(2n+1)/6, where n is the number of consecutive integers.
In this case, we have 4 consecutive integers, so n = 4. Substituting this into the formula, we get (4)(5)(9)/6 = 20.
To prove that this is the smallest possible value, we can use the concept of the arithmetic mean and quadratic mean inequality. The arithmetic mean of a set of numbers is always greater than or equal to the quadratic mean.
In this problem, the arithmetic mean is (a+b+c+d)/4, and the quadratic mean is √[(a^2 + b^2 + c^2 + d^2)/4]. Since we are trying to minimize the value of 4(a^2 + b^2 + c^2 + d^2) - (a+b+c+d)^2, we want to minimize the quadratic mean, which is √[(a^2 + b^2 + c^2 + d^2)/4].
Now, since a, b, c, and d are consecutive integers, we can write them as a = x, b = x+1, c = x+2, and d = x+3. Substituting these values into the quadratic mean, we get √[(x^2 + (x+1)^2 + (x+2)^2 + (x+3)^2)/4] = √[(6x^2 + 24x + 14)/4] = √(3x^2 + 12x + 7).
To minimize this expression, we can take the derivative and set it equal to 0. This gives us 6x + 12 = 0, or x = -2. Substituting this back into the expression, we get √(3(-2)^2 + 12(-2) + 7) = √(12 - 24 + 7) = √(-5) = undefined.
Since the quadratic mean cannot be negative, this means that the minimum value occurs when x = -2, which gives us
