Proving Inequalities Using Mathematical Induction

In mathematics, proving inequalities can be a powerful and elegant way to establish the validity of a statement for all natural numbers. One common and effective method to achieve this is through the process of mathematical induction. This article explains how to apply mathematical induction to prove inequalities, providing several examples and important considerations.

Example 1: Proving ( n^2 leq 2^n ) for All ( n in mathbb{N} )

Let's consider the inequality ( n^2 leq 2^n ). To prove this inequality for all natural numbers ( n ), we follow a two-step process: the base case and the inductive step.

Base Case

The base case is the smallest value of ( n ) for which we want the inequality to hold. In this example, we start with ( n 4 ).

When ( n 4 ), we have ( 4^2 16 ) and ( 2^4 16 ). Clearly, ( 16 leq 16 ), so the base case is true.

Inductive Step

For the inductive step, we assume that the inequality is true for some arbitrary natural number ( k geq 4 ), i.e., ( k^2 leq 2^k ). Next, we need to show that the inequality is also true for ( k 1 ).

We start by expressing ( (k 1)^2 ) in terms of ( k ):

[ (k 1)^2 k^2 2k 1 ]

Given our inductive hypothesis ( k^2 leq 2^k ), we need to compare ( k^2 2k 1 ) to ( 2^{k 1} ).

Since ( k^2 leq 2^k ), we have:

[ k^2 leq 2^k ]

Adding ( 2k 1 ) to both sides:

[ k^2 2k 1 leq 2^k 2k 1 ]

Now, note that ( 2k 1

[ 2^k 2k 1

Combining these inequalities, we get:

[ k^2 2k 1 leq 2^k 2k 1

Thus, ( (k 1)^2 leq 2^{k 1} ), completing the inductive step. Since the base case and inductive step are both true, the inequality ( n^2 leq 2^n ) holds for all ( n geq 4 ).

Example 2: Proving ( frac{1}{a^n} leq frac{1}{n a} ) for ( 1

Consider the inequality ( frac{1}{a^n} leq frac{1}{n a} ) for ( 1 Base Case: When ( n 1 ): (frac{1}{a} leq frac{1}{a}), which is clearly true. Inductive Step: Assume the inequality holds for ( n k ): (frac{1}{a^k} leq frac{1}{k a}). We need to show it holds for ( n k 1 ): (frac{1}{a^{k 1}} leq frac{1}{(k 1) a}). ( frac{1}{a^{k 1}} frac{1}{a^k cdot a} leq frac{1}{k a cdot a} frac{1}{k a^2} ) Since ( a 0 ), and thus ( k a^2 > (k 1) a ). Hence, (frac{1}{a^{k 1}} leq frac{1}{(k 1) a}).

The inequality is proved by induction for all ( n geq 1 ).

General Considerations in Proving Inequalities Using Induction

When proving inequalities using mathematical induction, there are several important considerations:

Base Case: Start with the smallest value of ( n ) and verify that the inequality holds. Inductive Hypothesis: Assume the inequality holds for ( n k ). Inductive Step: Prove the inequality for ( n k 1 ) based on the inductive hypothesis. Comparative Growth Rates: If the growth rates of the functions ( f(n) ) and ( g(n) ) are comparable, consider other methods such as analyzing ( f(n) - g(n) ) directly. Careful Verification: Especially when the inequality seems to be extremely sharp, check for large values to ensure the incorrectness of a false assumption.

By following these steps and considerations, you can effectively prove many inequalities using mathematical induction.