Proving Mathematical Induction for Inequalities Involving Averages

Mathematical induction is a powerful technique used to prove statements for all positive integers. In this article, we will delve into how to apply mathematical induction to prove an inequality involving averages, specifically focusing on the relationship between the arithmetic mean and the geometric mean of a sequence of positive numbers.

Introduction

Consider the problem of proving the inequality for the arithmetic mean (A) and the geometric mean (G) of a sequence of positive numbers a_1, a_2, ..., a_k:

[ frac{a_{1} a_{2} ... a_{k}}{k} geq sqrt[k]{a_{1}a_{2}...a_{k}} ]

This problem can be tackled using mathematical induction. The goal is to show that the inequality holds for a base case, and then to prove that if it holds for some positive integer k, it also holds for k 1.

Base Case

Let's first verify if the inequality holds for the base case, i.e., when k 1.

For the left-hand side, we have:

[ frac{a_1}{1} a_1 ]

For the right-hand side, we have:

[ a_{1}^{1/1} a_1 ]

Hence, the base case is trivially satisfied.

Induction Step

Assume the inequality holds for some positive integer k in mathbb{Z}^{ } (Induction Hypothesis):

[ frac{a_{1} a_{2} ... a_{k}}{k} geq sqrt[k]{a_{1}a_{2}...a_{k}} ]

We need to show that this implies the inequality holds for k 1 (i.e., for k_1 k 1):

[ frac{a_{1} a_{2} ... a_{k} a_{k 1}}{k 1} geq sqrt[k 1]{a_{1}a_{2}...a_{k}a_{k 1}} ]

Left Hand Side

The left-hand side can be rewritten as:

[ frac{a_{1} a_{2} ... a_{k}}{k} frac{a_{k 1}}{k 1} ]

Using the induction hypothesis, we can express this as:

[ frac{a_{1} a_{2} ... a_{k}}{k} frac{a_{k 1}}{k 1} geq sqrt[k]{a_{1}a_{2}...a_{k}} frac{a_{k 1}}{k 1} ]

Right Hand Side

The right-hand side can be expressed as:

[ sqrt[k 1]{a_{1}a_{2}...a_{k}a_{k 1}} (a_{1}a_{2}...a_{k}a_{k 1})^{1/(k 1)} ]

This can be further broken down as follows:

[ (a_{1}a_{2}...a_{k})^{1/(k 1)} (a_{k 1})^{1/(k 1)} sqrt[k]{a_{1}a_{2}...a_{k}} cdot (a_{k 1})^{1/(k 1)} ]

Show the Left Hand Side is Greater or Equal to the Right Hand Side

Therefore, we must show:

[ sqrt[k]{a_{1}a_{2}...a_{k}} frac{a_{k 1}}{k 1} geq sqrt[k]{a_{1}a_{2}...a_{k}} cdot (a_{k 1})^{1/(k 1)} ]

Define:

[ y frac{1}{k}a_{k 1} implies log y log a_{k 1} - log k leq log a_{k 1} ]

And:

[ z a_{1}a_{2}...a_{k} cdot (a_{k 1})^{1/k} implies log z log (a_{1}a_{2}...a_{k}) frac{1}{k}log (a_{k 1}) ]

Then:

[ log z log (a_{1}a_{2}...a_{k}) frac{1}{k}log (a_{k 1}) log (a_{1}a_{2}...a_{k}) log (a_{k 1})^{1/k} geq log (a_{k 1})^{1/k} geq log y ]

This shows:

[ z geq y implies a_{1}a_{2}...a_{k} cdot (a_{k 1})^{1/k} geq frac{1}{k}a_{k 1} ]

Therefore:

[ frac{a_{1} a_{2} ... a_{k} a_{k 1}}{k 1} geq (a_{1}a_{2}...a_{k}a_{k 1})^{1/(k 1)} ]

Proven the inequality for the case of k 1.

Conclusion

This step-by-step proof by mathematical induction confirms the inequality for all positive integers k. Therefore, we have successfully shown the inequality holds for both the base case and the induction step, thus the inequality is proven for all positive integers k.