# Mathematical Induction Inequality Proofs

Mathematical Induction Inequality is being used for proving inequalities. It is quite often applied for the subtraction and/or greatness, using the assumption at the step 2. Let’s take a look at the following hand-picked examples.

## Practice Questions for Mathematical Induction Inequality

### Basic Mathematical Induction Inequality

Prove \( 4^{n-1} \gt n^2 \) for \(n \ge 3\) by mathematical induction.

Step 1: Show it is true for \( n=3 \).

LHS \(=4^{3-1} = 16 \)

RHS \(=3^2=9 \)

LHS > RHS

Therefore it is true for \( n=3 \).

Step 2: Assume that it is true for \( n=k \).

That is, \( 4^{k-1} > k^2 \).

Step 3: Show it is true for \( n=k+1 \).

That is, \( 4^{k} > (k+1)^2 \).

\( \begin{aligned} \displaystyle \require{color}

\text{LHS } &= 4^k \\

&= 4^{k-1+1} \\

&= 4^{k-1} \times 4 \\

&\gt k^2 \times 4 &\color{red} \text{by the assumption } 4^{k-1} > k^2 \\

&= k^2 + 2k^2 + k^2 &\color{red} 2k^2 > 2k \text{ and } k^2 > 1 \text{ for } k \ge 3 \\

&\gt k^2 + 2k + 1 \\

&= (k+1)^2 \\

&=\text{RHS} \\

\text{LHS } &\gt \text{ RHS}

\end{aligned} \)

Therefore it is true for \( n=k+1 \) assuming that it is true for \( n=k \).

Therefore \( 4^{n-1} \gt n^2 \) is true for \( n \ge 3 \).

### Mathematical Induction Inequality using the Difference

It is quite often used to prove \( A > B \) by \( A-B >0 \).

Prove \( n^2 \lt 2^n \) for \( n \ge 5 \) by mathematical induction.

Step 1: Show it is true for \( n=5 \).

LHS \( = 5^2 = 25 \)

RHS \( = 2^5 = 32 \)

LHS \( \lt \) RHS

It is true for \( n=5 \).

Step 2: Assume that it is true for \( n=k \).

That is, \( k^2 \lt 2^k \).

Step 3: Show it is true for \( n=k+1 \).

That is, \( (k+1)^2 \lt 2^{k+1}. \)

\( \begin{aligned} \displaystyle \require{color}

\text{RHS } – \text{ LHS } &= 2^{k+1} – (k+1)^2 \\

&= 2 \times 2^k – (k^2+2k+1) \\

&\gt 2 \times k^2 – (k^2+2k+1) &\color{red} \text{ by the assumption from Step 2} \\

&= k^2 -2k -1 \\

&= (k-1)^2 -2 \\

&\gt 0 &\color{red} \text{, since } k \ge 5 \text{ and so } (k-1)^2 \ge 16 \\

2^{k+1} – (k+1)^2 &\gt 0 \\

(k+1)^2 &\lt 2^{k+1} \\

\end{aligned} \)

Therefore it is true for \( n=k+1 \) assuming it is true for \( n=k \).

Therefore it is true for \( n=k+1 \) is true for \( n \ge 5 \).

### Related Topics

**Best Examples of Mathematical Induction Divisibility**

**Mathematical Induction Fundamentals**

**Mathematical Induction Inequality Proof with Factorials**

**Mathematical Induction Inequality Proof with Two Initials**