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Prove the arithmetic mean – geometric mean inequality

by RoRi

We define the geometric mean G of real numbers x_1, \ldots, x_n by

    \[ G := (x_1 \cdot \cdots \cdot x_n)^{1/n}. \]

We also recall the definition of the pth power mean, M_p:

    \[ M_p = \left( \frac{x_1^p + \cdots + x_n^p}{n} \right)^{1/p}. \]

  1. Prove the arithmetic mean – geometric mean inequality, i.e., prove G \leq M_1 where M_1 is the pth power mean with p=1 (also known as the arithmetic mean).
  2. For integers p,q with q < 0 < p, prove that M_q < G < M_p if x_1, \ldots, x_n are not all equal.
  1. Proof. First, if x_1 = \cdots = x_n, then

        \[ G = (x_1 \cdot \cdots \cdot x_n)^{1/n} = (x_1^n)^{1/n} = x_1 \qquad \text{and} \qquad M_1 = \left( \frac{x_1 + \cdots + x_n}{n} \right) = \left( \frac{nx_1}{n} \right) = x_1. \]

    Hence, G = M_1 for the case x_1 = \cdots = x_n.
    Now, if x_1, \ldots, x_n are not all equal then first we write,

        \[ G^n = \left( (x_1 \cdot \cdots \cdot x_n)^{1/n}\right)^n = x_1 \cdot \cdots \cdot x_n. \]

    Then using the previous exercise, we proceed,

        \begin{align*} &&\left( \frac{1}{G^n} \right) (x_1 \cdot \cdots \cdot x_n) &= 1 \\ \implies && \left( \frac{x_1}{G} \right) \left( \frac{x_2}{G} \right) \cdot \cdots \cdot \left( \frac{x_n}{G} \right) &= 1 \\ \implies && \frac{x_1}{G} + \frac{x_2}{G} + \cdots + \frac{x_n}{G} &> n & (\text{previous exercise})\\ \implies && x_1 + x_2 + \cdots + x_n &> n \cdot G \\ \implies && M_1 &> G. \end{align*}

    Therefore, if x_1, \ldots, x_n are not all equal then we have the strict inequality G < M_1, as requested. \qquad \blacksquare

  2. Proof. We’ll start with the inequality on the right first. So, we want to show G < M_p for any n positive real numbers x_1, \ldots, x_n not all equal, where p is a positive integer. First, we’ll want to observe that if x_1, \ldots, x_n are positive real numbers, not all equal, then x_1^p, \ldots, x_n^p are also positive real numbers, not all equal. So from the definition of M_p and letting M_p (x_1^p, \ldots, x_n^p) denote the pth power mean of the numbers x_1^p, \ldots, x_n^p, we have

        \begin{align*} M_1 (x_1^p, \ldots, x_n^p) &= \left( \frac{x_1^p + \cdots + x_n^p}{n} \right) \\ &= \left( \left( \frac{x_1^p + \cdots + x_n^p}{n} \right)^{1/p} \right)^p \\ &= (M_p(x_1, \ldots, x_n))^p. \end{align*}

    So, the observation is that (M_p (x_1, \ldots, x_n))^p = M_1(x_1^p, \ldots, x_n^p). Now, using part (a), we have,

        \begin{align*} G(x_1, \ldots, x_n)^p &= (x_1 \cdots x_n)^{p/n}\\ &= (x_1^p \cdots x_n^p)^{1/n} \\ &= G(x_1^p, \ldots, x_n^p) \\ &< M_1 (x_1^p, \ldots, x_n^p) & (\text{part (a)})\\ &= (M_p (x_1, \ldots, x_n))^p. \end{align*}

    Hence, for any positive real numbers x_1, \ldots, x_n, not all equal we have (G(x_1, \ldots, x_n))^p < (M_p (x_1, \ldots, x_n))^p which implies (see this exercise) G < M_p. This gives us the inequality on the right.
    Now, for q a negative integer, we must show M_q < G. Since q < 0 we know that -q>0 and so G < M_{-q} from the inequality we just proved. So,

        \[ G^{-q} < (M_{-q})^{-q} \quad \implies \quad G^q > M_q^q \quad \implies \quad G > M_q. \]

    Where we have used the same exercise again, and the fact that G^{-q} = \frac{1}{G^q}. \qquad \blacksquare

3 comments

  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

        \[(G(x_1, \ldots, x_n))^p < (M_p (x_1, \ldots, x_n))^p\]

    . The inequality implies the product not the other way around. Is it not that simple as taking p-squared to both sides?

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