A set of n > 3 real numbers has sum at least n and the sum of the squares of the numbers is at least n2. Show that the largest positive number is at least 2.
Solution
Let the numbers be x1, x2, ... , xn. Notice first that x1 = x2 = ... = xn-1 = 2, xn = 2 - n, gives ∑ xi = (n - 1)2 + (2 - n) = n, ∑ xi2 = (n - 1)4 + (4 - 4n + n2) = n2, so the inequality is best possible.
Suppose the result is false. So we have a set of numbers with S xi ≥ n, ∑ xi2 ≥ n2 and max xi < 2. At least one of the numbers must be negative, since otherwise we have n ≥ 4, so n2 ≥ 4n > ∑ xi2. Contradiction. This allows us to assume that ∑ xi = n, for if it is greater, we may just decrease a negative xi until it becomes true (∑ xi2 will be increased, so it will remain at least n2).
Now suppose two of the xi, namely x and y, are less than 2. Then if we replace them by 2 and x + y - 2, the sum is unaffected and the sum of squares is increased by 2(2 - x)(2 - y). Since we start with all the xi less than 2, we may do this repeatedly until we reach a set with all the numbers 2 except one. Since the sum is unchanged, the other number must be 2 - n, and, as shown above, that makes the sum of the squares n2. But we have increased the sum of the squares at each step. Contradiction.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002