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# Find a value which minimizes a given sum

Let . Prove

is minimal when is the arithmetic mean of .

Proof. Recall that the arithmetic mean of is given by

Now, consider the given sum

To find the minimum we first take the derivative, using the linearity of the derivative over this (finite) sum, (i.e., using that ),

Setting this equal to 0 and solving,

# Prove that different functions may have the same average

Let be a continuous, strictly monotonic function on with inverse , and let be given positive real numbers. Then define,

This is called the mean of with respect to . (When for , this coincides with the th power mean from this exercise).

Show that if with , then .

Proof. Let with . Then, has an inverse since it is strictly monotonic (since it is the composition of and the linear function , both of which are strictly monotonic for ). Its inverse is given by

So,

# Show the mean of a strictly monotonic function lies in an interval

Let be a continuous, strictly monotonic function on with inverse , and let be given positive real numbers. Then define,

This is called the mean of with respect to . (When for , this coincides with the th power mean from this exercise).

Show that

Proof. Since is strictly monotonic on the positive real axis and are positive reals, we know is strictly increasing or strictly decreasing, and correspondingly we have,

First, assume is striclty increasing, then

Since is strictly increasing so is its inverse (by Apostol’s Theorem 3.10); thus, we have

If is strictly decreasing then

# Prove connection between power mean and arithmetic mean of a function

Let be a continuous, strictly monotonic function on with inverse , and let be given positive real numbers. Then define,

This is called the mean of with respect to . (When for , this coincides with the th power mean from this exercise).

Show that

Proof. Since is the inverse of we know for all in the range of , i.e., for all such that there is some such that .

By the definition of then, we have that

So, if is in the domain of then we are done. Since is the inverse of it’s domain is equal to the range of . We show that this value is in the range of using the intermediate value theorem.

Without loss of generality, assume is strictly increasing (the alternative assumption, that is strictly decreasing will produce an almost identical argument). Then, since are all positive real numbers we have . (Here if we’d assumed that was strictly decreasing the roles inequalities would be reversed.) Then we have,

Hence, by the intermediate value theorem, since

there must be some such that

Thus, is in the domain of , so

# Prove the arithmetic mean – geometric mean inequality

We define the geometric mean of real numbers by

We also recall the definition of the th power mean, :

1. Prove the arithmetic mean – geometric mean inequality, i.e., prove where is the th power mean with (also known as the arithmetic mean).
2. For integers with , prove that if are not all equal.

1. Proof. First, if , then

Hence, for the case .
Now, if are not all equal then first we write,

Then using the previous exercise, we proceed,

Therefore, if are not all equal then we have the strict inequality , as requested

2. Proof. We’ll start with the inequality on the right first. So, we want to show for any positive real numbers not all equal, where is a positive integer. First, we’ll want to observe that if are positive real numbers, not all equal, then are also positive real numbers, not all equal. So from the definition of and letting denote the th power mean of the numbers , we have

So, the observation is that . Now, using part (a), we have,

Hence, for any positive real numbers , not all equal we have which implies (see this exercise) . This gives us the inequality on the right.
Now, for a negative integer, we must show . Since we know that and so from the inequality we just proved. So,

Where we have used the same exercise again, and the fact that