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# 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 1. fran says: , not . Please correct it or it’s cheating.

2. Daniel says:

For (b), you just needed to write the AG-GM inequality for $x_{1}^p, … , x_{n}^p$ and raise the inequality to $1/p$. You’re welcome.

3. Anonymous says:

I don´t understand why you link to the other exercise in . The inequality implies the product not the other way around. Is it not that simple as taking p-squared to both sides?