Prove that at least one of the real number a1,a2,..,an is greater than or equal to the average of these number

Question

use methods of proof

Answer 1

Let A be the average of a1,a2,...,an. That is,

A = (a1+a2+...+an) / n.

Suppose a1
Adding all of these inequalities, you get

(a1+a2+...+an) < nA,

so

(a1+a2+...+an)/n < A

or

A < A

which is a contradiction. Therefore at least one of the numbers must be >= A.

Answer 2

Let X be the average of the real numbers {a1,a2,...,aN}, then, X = (a1 + a2 + ... + aN)/N

Therefore,
X * N = a1 + a2 + ... + aN. (equation 1)

Now, to complete the proof by contradiction, assume that X is greater than each and every element of {a1, a2, ..., aN}.

Since X is strictly greater than a1, if I subtract X from the left side of equation 1 and I subtract a1 from the right side of equation 1, I will have:

X(N-1) < a2 + a3 + ... + aN

Again, subtract X from the left and subtract a2 from the right. As I am subtracting a larger number from the left side, the inequality will continue:

X(N-2)< a3 + ... + aN.

Continue doing this until you have subtracted X from the left side a total of N times and you have subtracted each of the a(i) from the right. You will then have:
X(N-N) < 0
or
0<0.

Since 0 is not strictly less than 0, it must follow that our assumption was in error. Our assumption was that no element of {a1, a2, ... , aN} was greater than or equal to X. Therefore, there must exist some a(i) in the given set of real numbers that must satisfy the condition that a(i) >= X.

Answer 3

Assume all are < average. The average of n numbers Sum S is S/number =A. Sum of n each < A is less than S. Therefore at least one must be larger than A.