Examples – Proof by Mathematical Induction for Inequalities

Lata Agarwal Mathematical Induction, Mathpedia
mathematical induction, prove inequalities using mathematical induction examples with solution, divisibility statements, summation identities

Prove Inequalities using Mathematical Induction examples with solutions

Question 1:

Prove the following inequality by the principle of mathematical induction:

{1^2} + {2^2} + \ldots + {n^2} > \frac{{{n^3}}}{3}, n ∈ N.

Solution:

We can write,

P\left( n \right):\quad {1^2} + {2^2} + \ldots + {n^2} > \frac{{{n^3}}}{3}

Step 1:

We first show that the basis for induction P(1) is true, that is, P(n) is true for n = 1.

Left Side = {1^2} = 1,

Right Side = \frac{{{1^3}}}{3} = \frac{1}{3}.

Since, 1 > \frac{1}{3}, that is, Left Side > Right Side

Hence, P(1) is true, that is, P(n) is true for n = 1.

Step 2:

Assume that P(k) is true for some natural number k, that is,

{1^2} + {2^2} + \ldots + {k^2} > \frac{{{k^3}}}{3}

We need to prove that P(k +1) is also true.

Because P(k) is true, so we have,

{1^2} + {2^2} + \ldots + {k^2} > \frac{{{k^3}}}{3}

Add {\left( {k + 1} \right)^2} to both sides:

{1^2} + {2^2} + \ldots + {k^2} + {\left( {k + 1} \right)^2}

> \frac{{{k^3}}}{3} + {\left( {k + 1} \right)^2}

= \frac{{{k^3} + 3{{\left( {k + 1} \right)}^2}}}{3}

= \frac{{{k^3} + 3\left( {{k^2} + 2k + 1} \right)}}{3}

= \frac{{{k^3} + 3{k^2} + 6k + 3}}{3}

= \frac{{{{\left( {k + 1} \right)}^3}}}{3}.

That is, we get

{1^2} + {2^2} + \ldots + {k^2} + {\left( {k + 1} \right)^2} > \frac{{{{\left( {k + 1} \right)}^3}}}{3}.

Thus, P(k +1) is true whenever P(k) is true.

Therefore, from the principle of mathematical induction, the statement:

{1^2} + {2^2} + \ldots + {n^2} > \frac{{{n^3}}}{3}

is true for all natural number n.

Question 2:

Prove the following statement for all integers n \ge 4 by the principle of mathematical induction:

n! > {n^2}.

Solution:

We can write,

P\left( n \right):\quad n! > {n^2}

Step 1:

We first show that the basis for induction P(4) is true, that is, P(n) is true for n = 4.

Left Side = 4! = 4 \times 3 \times 2 \times 1 = 24,

Right Side = {4^2} = 16.

Since, 24 > 16, that is, 4! > {4^2}, that is, Left Side > Right Side

Hence, P(4) is true, that is, P(n) is true for n = 4.

Step 2:

Assume that P(k) is true for all integers k \ge 4, that is,

k! > {k^2}

We need to prove that P(k +1) is also true, that is, we need to prove that \left( {k + 1} \right)! > {\left( {k + 1} \right)^2} if k! > {k^2}.

To prove required from this, consider,

\left( {k + 1} \right)! = \left( {k + 1} \right) \cdot k!

> \left( {k + 1} \right) \cdot {k^2}                  (Apply k! > {k^2} by the induction hypothesis)

> 4{k^2}                               (Because k \ge 4 so k + 1 > 4) 

= {k^2} + 2{k^2} + {k^2}

> {k^2} + 2k + 1

= {\left( {k + 1} \right)^2}.

That is, we get

\left( {k + 1} \right)! > {\left( {k + 1} \right)^2}.

Thus, P(k +1) is true whenever P(k) is true for all integers k \ge 4.

Therefore, from the principle of mathematical induction, the statement:

n! > {n^2}

is true for all integers n \ge 4.

Tags: mathematical induction examples with solutions, prove inequalities by induction

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About Lata Agarwal 270 Articles
M.Phil in Mathematics, skilled in MS Office, MathType, Ti-83, Internet, etc., and Teaching with strong education professional. Passionate teacher and loves math. Worked as a Assistant Professor for BBA, BCA, BSC(CS & IT), BE, etc. Also, experienced SME (Mathematics) with a demonstrated history of working in the internet industry. Provide the well explained detailed solutions in step-by-step format for different branches of US mathematics textbooks.
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