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How to prove Bernoulli’s inequality with mathematical induction?

Bernoulli's inequality with mathematical induction


Mathematical induction is no strange to mathematics students. It is a technique to prove any formula, equation, or theorem. It can be used to prove that identity is valid. On the other hand, Bernoulli’s inequality is used in real analysis. It is an inequality that approximates the exponentiation of 1+x.

We can use this technique for theoretical purposes as well as real-life hypotheses and conclusions. By using the correct data and mathematical induction, you can conclude that how the prices of real estate will escalate or remain in the same range.

In this post, we will use mathematical induction to prove Bernoulli’s inequality for a function. But before doing so, we will explain mathematical induction and Bernoulli’s inequality. Keep on reading to find out solved examples of Bernoulli’s inequality for various functions.

What is Bernoulli’s inequality?

Jacob Bernoulli was a famous mathematician known for his work in Calculus and Probability. He also discovered the constant e. The Bernoulli’s inequality is named after him. He first came up with the concept of inequality in 1689.

Bernoulli’s inequality is an inequality that estimates exponentiations of 1 + x. It is mostly employed in real life predictions analysis.

(1 + x)r ≥ 1 + rx
In this expression, x represents the real numbers and x ≥ -1, while r represents the real number and r ≠ 0.

Many classical inequalities based on the theory of inequalities not only facilitate the advancement of the inequality theory but also lead to many applications in pure and applied mathematics. Bernoulli’s inequality is one of the most well-known disparities in mathematics.

What is mathematical induction?

Mathematical induction is a common and effective proof technique. In its heart, it is an appeal to an intuitive notion. Induction proofs are often used in computer science to demonstrate that an algorithm functions as expected and runs in a certain period of time.

It is also used for mathematical proofing. It is a method for verifying output or establishing expressions for natural numbers.

The procedure uses the following steps to prove an expression or statement.

  1. Base case: It attests that an expression is true for the initial value.
  2. Inductive case: It verifies that if the expression or statement is true for the nth iteration, then it is also true for (n+1) th iteration or n+1.

The theory of complete induction holds true not only for statements about natural numbers but also for statements about elements of any well-founded set. The theory poses that a set with an irreflexive relation is less than a set with no infinite descending chains. Any set of cardinal numbers which includes the set of natural numbers is well-founded.

Principle of Mathematical Induction

To understand the principle of mathematical induction, we will assume A as a set of natural numbers.

Let A be a set of natural numbers such that it holds the two properties given below.

1 ∈ A, and

For every natural number n if n A then n + 1 A

Then, A = N = {1, 2, . . .}

Which means, A contains all natural numbers.

Now that you understand the basic concepts of both terms discussed above, it’s time to do the real work.

Let’s prove Bernoulli’s inequality for a few functions

There are two ways to prove the inequality of Bernoulli’s equation. The first one is, using a mathematical induction calculator. An online calculator can facilitate your calculations and you may also get the steps of the calculation.

The second method is using the technique of mathematical induction. In this method, we will deal with equations and substitutions to get to the proof for a given function. You can opt-out for any method depending on your situation.

Here are a few examples to prove Bernoulli’s inequality with mathematical induction.

Prove: 1 + 3 + 5 + . . . + (2n−1) = n2

Step 1: Check if the given function is true for n = 1.

1 = 12  True

Step 2: Let’s assume the function is true for n = k.

1 + 3 + 5 + . . . + (2k − 1) = k2  True

Step 3: Now prove that the function is true for “k + 1“.

1 + 3 + 5 + . . . + (2k − 1) + (2(k + 1) −1) = (k + 1)2

We already know that,

1 + 3 + 5 + . . . + (2k − 1) = k2

Except for the last term in the expression, we can do a substitution here.

k2 + (2(k+1) −1) = (k+1)2

Step 4: In this step, expand every term in the above equation.

k2 + 2k + 2 − 1 = k2 + 2k+1

Simplify the equation to get,

k2 + 2k + 1 = k2 + 2k + 1

If both sides are identical, then it is true.

1 + 3 + 5 + … + (2(k+1) −1) = (k+1)2  True

The technique we used to prove the Bernoulli’s inequality for the function 1 + 3 + 5 + . . . + (2n−1) = n2 is mathematical induction. Using this method, we can find proof of inequality for any function. If you are performing a real analysis, try reaching out to an expert so that you don’t make any mistakes.

The above proof is performed on a simple function. The calculations may get complicated when using a multi-variable function or a complex one. To avoid getting stuck in the process, we recommend you get a good grip on this concept and start using simple functions in the beginning.


All in all, now we are sure that Bernoulli’s inequality can be proved with mathematical induction regardless of the type of function.

There a few things to consider. You should follow both cases of mathematical induction while proving an expression or statement. Follow all steps we have mentioned in the above example. Moreover, the comparison at the end is inevitable to find out if the statement is true or false.


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