A Proof of the Arithmetic Mean–Geometric Mean Inequality

Branko Ćurgus

By \(\mathbb{N}\) we denote the set of positive integers and by \(\mathbb{R}_{\gt 0}\) we denote the set of all positive real numbers.

Lemma 1.

For every \( m\in \mathbb{N}\) and all \(x \in \mathbb{R}\) we have \[ m\mkern1mu x^{m+1} - (m + 1)\mkern1mu x^m + 1 = (x - 1)^2 \sum_{k=1}^{m} k \mkern 1mu x^{k-1}. \]

Proof.

Let \(m\in\mathbb{N}\), \(k\in \{0\}\cup \mathbb{N}\) and \(x\in\mathbb{R}\) be arbitrary. Set \[ X_k = k\mkern1mu x^{k+1} - (k + 1)\mkern1mu x^k + 1. \] For all \(k\in\{1,\ldots,m\}\) we have \[ \begin{aligned} X_{k} - X_{k-1} & = \bigl( k\mkern1mu x^{k+1} - (k + 1)\mkern1mu x^k + 1 \bigr) - \bigl( (k-1)\mkern1mu x^{k} - k\mkern1mu x^{k-1} + 1 \bigr) \\ & = k\mkern1mu x^{k+1} - 2k\mkern1mu x^{k} + k\mkern1mu x^{k-1} \\ & = (x^2 -2 x + 1) \mkern1.5mu k \mkern1mu x^{k-1} \\ & = (x-1)^2 \mkern1.5mu k \mkern1mu x^{k-1}. \end{aligned} \] Since \(X_0 = 0\), we have \[ m\mkern1mu x^{m+1} - (m + 1)\mkern1mu x^m + 1 = \sum_{k=1}^{m} \bigl(X_{k} - X_{k-1} \bigr) = (x - 1)^2 \sum_{k=1}^{m} k \mkern 1mu x^{k-1}.\,\square \]

Lemma 2.

For all \(m\in\mathbb{N}\) and all \(s \in \mathbb{R}_{\gt 0}\) we have \(m\mkern1mu s^{-1/m} + s\mkern1mu \geq m + 1\) with equality if and only if \(s=1.\)

Proof.

We apply Lemma 1 with \(x = s^{-1/m}\) to establish the identity \[ \begin{aligned} m\mkern1mu s^{-1/m} + s - (m + 1) & = s\mkern1mu \Bigl( m\mkern1mu s^{-1-1/m} - (m + 1)\mkern1mu s^{-1} + 1 \Bigr) \\ & = s\mkern1mu \Bigl( m\mkern1mu \bigl( s^{-1/m} \bigr)^{m+1} - (m + 1)\mkern1mu \bigl(s^{-1/m}\bigr)^m + 1 \Bigr) \\ & = s\mkern1mu \bigl( s^{-1/m} - 1 \bigr)^2 \sum_{k=1}^{m} k \mkern 1mu s^{-(k-1)/m}, \end{aligned} \] which yields the claim, since \(s \gt 0\) and \(\sum_{k=1}^{m} k\mkern1mu \mkern 1mu s^{-(k-1)/m} \gt 0.\) \(\square\)

Theorem (AM–GM).

For every \(n\in\mathbb{N}\) the following statement holds:

For all \(n\)-tuples \((a_1,\ldots,a_n) \in (\mathbb{R}_{\gt 0})^n\) we have \[ \bigl(a_1 \cdots a_n \bigr)^{1/n} \leq \frac{1}{n} \bigl(a_1 + \cdots + a_n \bigr). \]

Proof.

We proceed with a proof by Mathematical Induction. For \(n\in\mathbb{N}\) denote the claim of the theorem by \(\operatorname{AMGM}(n).\) Clearly \(\operatorname{AMGM}(1)\) is true. Let \(m\in\mathbb{N}\) be arbitrary and assume \(\operatorname{AMGM}(m).\) For \((x_1,\ldots,x_m,x_{m+1}) \in (\mathbb{R}_{\gt 0})^{m+1}\) let \(G = (x_1\cdots x_{m+1})^{1/(m+1)}\) be its geometric mean. Then \(\prod_{k=1}^{m+1} ( x_k/G ) = 1\), hence \(\prod_{k=2}^{m+1} ( x_k/G ) = G/x_1.\) By the inductive hypothesis \(\operatorname{AMGM}(m)\) we have \[ \frac{1}{m} \sum_{k=2}^{m+1} ( x_k/G ) \geq ( G/x_1 )^{1/m}. \tag{1} \] Factoring \(G\), then using (1), and finally Lemma 2 with \(s = x_1/G \gt 0\), we obtain \[ \begin{aligned} \sum_{k=1}^{m+1} x_k & = G \left( x_1/G + \sum_{k=2}^{m+1} ( x_k/G ) \right) \\ & \geq G \left( x_1/G + m ( G/x_1 )^{1/m} \right) \\ & \geq G\,(m+1), \end{aligned} \] proving \(\operatorname{AMGM}(m+1)\) and completing the proof by Mathematical Induction. \(\square\)