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4.4: Arithmetic-Geometric Inequality

  • Page ID
    • Bob Dumas and John E. McCarthy
    • University of Washington and Washington University in St. Louis
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    We have presented modest generalizations of basic mathematical induction (Corollary \(4.8\) and Corollary 4.9). The formality of our approach might suggest that induction is a rigid technique that must be applied inflexibly in a specific prescriptive way. To a mathematician induction is governed by two ideas:

    (1) Induction uses the well ordering of the natural numbers, or more generally any well-ordered set, to prove universal statements quantified over the set.

    (2) Every element in the set over which you quantify must be accounted for by the induction.

    The formal characterizations of induction in Section \(4.2\) are sufficient but not necessary to achieve the objectives of a proof by induction. The theorem in this section will give you a sense about how the technique of induction can be extended. DEFINITION. Arithmetic mean Let \(a_{1}, \ldots, a_{N}\) be real numbers. The arithmetic mean of \(a_{1}, \ldots, a_{N}\) is \[\frac{1}{N}\left(\sum_{n=1}^{N} a_{n}\right) .\] Definition. Geometric mean Let \(a_{1}, \ldots, a_{N}\) be positive real numbers. The geometric mean of \(a_{1}, \ldots, a_{N}\) is \[\sqrt[N]{a_{1} \cdots a_{N}} .\] THEOREM 4.16. Arithmetic-geometric mean inequality Let \(a_{1}, \ldots, a_{N} \in\) \(\mathbb{R}^{+}\). Then \[\sqrt[N]{a_{1} \cdots a_{n}} \leq \frac{1}{N}\left(\sum_{n=1}^{N} a_{n}\right) .\] DisCUSSION. We prove this with an interesting argument due originally to Cauchy; our treatment is from the book [1]. We argue by induction on the size of the sample over which we are computing the means. After arguing the base case we show that if the inequality holds for the arithmetic and geometric means of \(N\) numbers, it necessarily holds for the means of \(2 N\) numbers. This implies that the theorem holds for the means of \(2^{N}\) numbers for any \(N \in \mathbb{N}\) (by a standard induction argument).

    We then show that the result holding for \(N\) numbers implies that it holds for \(N-1\) numbers. This implies that if the result holds at a natural number \(N\), the inequality holds for all means of fewer than \(N\) numbers. Given any \(k \in \mathbb{N}, 2^{k}>k\) and since the theorem holds for means of \(2^{k}\) numbers, it holds for means of \(k\) terms.

    Proof. We argue by induction on the number of terms on each side of the inequality.

    Base case: \((N=2)\)

    Let \(a_{1}, a_{2} \in \mathbb{R}^{+}\). Then \[\left(a_{1}-a_{2}\right)^{2}=a_{1}^{2}-2 a_{1} a_{2}+a_{2}^{2} \geq 0 .\] Therefore \[2 a_{1} a_{2} \leq a_{1}^{2}+a_{2}^{2},\] and \[\begin{aligned} 4 a_{1} a_{2} & \leq a_{1}^{2}+2 a_{1} a_{2}+a_{2}^{2} \\ &=\left(a_{1}+a_{2}\right)^{2} . \end{aligned}\] Thus \[2 \sqrt{a_{1} a_{2}} \leq a_{1}+a_{2} .\] Therefore the inequality holds for two terms.

    Induction step:

    Let \(P(N)\) be the statement that (4.17) holds for all \(a_{1}, \ldots, a_{N}>0\). We show that \(P(N) \Rightarrow P(2 N)\). Let \[G_{N}=\prod_{n=1}^{N} a_{n}\] and \[A_{N}=\left(\frac{\sum_{n=1}^{N} a_{n}}{N}\right) .\] So \[\begin{aligned} G_{2 N} &=\prod_{n=1}^{2 N} a_{n} \\ &=\left(\prod_{n=1}^{N} a_{n}\right)\left(\prod_{n=N+1}^{2 N} a_{n}\right) \\ & \leq_{I H}\left(\sum_{n=1}^{N} \frac{a_{n}}{N}\right)^{N}\left(\sum_{n=N+1}^{2 N} \frac{a_{n}}{N}\right)^{N} \end{aligned}\] Let \[B=\sum_{n=N+1}^{2 N} \frac{a_{n}}{N} .\] By the base case \[\begin{aligned} A_{N} B & \leq\left(\frac{A_{N}+B}{2}\right)^{2} \\ &=\left(A_{2 N}\right)^{2} \end{aligned}\] So \[\begin{aligned} \left(A_{N}\right)^{N} B^{N} &=\left(A_{N} B\right)^{N} \\ & \leq\left(\left(A_{2 N}\right)^{2}\right)^{N} \\ &=\left(A_{2 N}\right)^{2 N} \end{aligned}\] Thus \[G_{2 N} \leq\left(A_{2 N}\right)^{2 N} .\] Therefore, for any \(N \in \mathbb{N}^{+}\), \[P(N) \Rightarrow P(2 N) .\] Discussion. Let \(Q(N)\) be the statement \(P\left(2^{N}\right)\). Then the argument thus far is a standard proof by induction of \(\left(\forall N \in \mathbb{N}^{+}\right) Q(N)\). Of course we wish to show \((\forall N \in \mathbb{N}) P(N)\). We do this by proving \[\left(\forall N \in \mathbb{N}^{+}\right) P(N+1) \Rightarrow P(N) .\] Let \(N>2\). We prove that \[P(N+1) \Rightarrow P(N) .\] Assume \(P(N+1)\). Then \[\left(G_{N}\right)\left(A_{N}\right) \leq\left(\frac{\left(\sum_{n=1}^{N} a_{n}\right)+A_{N}}{N+1}\right)^{N+1} .\] DiscuSsion. Recall that \(G_{N}\) is the product of \(a_{1}, \ldots, a_{N}\). We are treating the sum \(A_{N}\) as the \(N+1^{\text {st }}\) factor, \(a_{N+1}\), and applying the inequality \(P(N+1)\).

    As \[\begin{aligned} \left(\frac{\left(\sum_{n=1}^{N} a_{n}\right)+A_{N}}{N+1}\right)^{N+1} &=\left(\frac{N A_{N}+A_{N}}{N+1}\right)^{N+1} \\ &=\left(A_{N}\right)^{N+1} \end{aligned}\] Inequality \(4.18\) gives \[G_{N} A_{N} \leq A_{N}^{N+1}\] and so \[G_{N} \leq\left(A_{N}\right)^{N}\] which is the statement \(P(N)\). So \[\left(\forall N \in \mathbb{N}^{+}\right) P(N+1) \Rightarrow P(N) .\] Hence for all \(N \geq 2, P(N)\).

    The arithmetic mean and geometric mean are different ways of understanding averages. They are related by the arithmetic geometric mean inequality (called the AGM inequality). Can we apply the inequality? Let’s consider an easy geometrical application of the case \(N=2\). Consider the rectangle with sides length \(a\) and \(b\). The perimeter of the rectangle is \[P=2 a+2 b\] and the area is \[A=a b .\] image

    In calculus you proved that the rectangle of fixed perimeter with the greatest area is the square. This can also be proved directly from the AGM inequality: \[\begin{aligned} P &=2 a+2 b \\ &=\frac{4 a+4 b}{2} \\ & \geq \sqrt{16 a b} \\ &=4 \sqrt{a b} . \end{aligned}\] So \[\frac{P^{2}}{16} \geq a b=A .\] Recall that \(P\) is fixed, and therefore so is \(\frac{P^{2}}{16}\), and we have shown that this is an upper bound for the area of the rectangle.

    Is this upper bound achieved? The area \(A\) of the rectangle varies according to the dimensions of the rectangle and if \(a=b\) \[\frac{P^{2}}{16}=\frac{(4 a)^{2}}{16}=A .\] Thus the maximum area of the rectangle is achieved when \(a=b\). This result can be generalized to higher dimensions - without the need for multivariable calculus.

    Proving theorems is not just a question of technique, though this must be mastered. It also requires creativity and insight. A beautiful collection of proofs is contained in the book [1] by Martin Aigner and Günter Ziegler.

    This page titled 4.4: Arithmetic-Geometric Inequality is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Bob Dumas and John E. McCarthy via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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