Mathematical Induction Inequality

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

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 Differences

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

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

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 \( n \ge 5 \).

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