Mathematical Induction Inequality Proof with Factorials uses one of the properties of factorials, \( n! = n(n-1)! = n(n-1)(n-2)! \).

# Worked Example of Mathematical Induction Inequality Proof with Factorials

Prove that \( (2n)! > 2^n (n!)^2 \) using mathematical induction for \(n \ge 2\).

Step 1: Show it is true for \( n =2 \).

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

\text{LHS } &= (2 \times 2)! = 16 \\

\text{RHS } &= 2^2 \times (2!) = 8 \\

\text{LHS } &> { RHS} \\

\end{aligned} \)

\( \therefore \text{It is true for } n =2 \)

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

Step 3: Show it is true for \(n =k+1\), that is \( (2k+2)! > 2^{k+1} \big[(k+1)!\big]^2. \)

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

\text{LHS } &= (2k+2)! \\

&= (2k+2)(2k+1)(2k)! \\

&= 2(k+1)(2k+1)(2k)! \\

&> 2(k+1)(2k+1)2^k (k!)^2 &\color{red} \text{by assumption} \\

&> (k+1)(2k+1)2^{k+1} (k!)^2 \\

&> (k+1)(k+1)2^{k+1} (k!)^2 &\color{red} 2k+1>k+1 \\

&= 2^{k+1} \big[(k+1)k!\big]^2 \\

&= 2^{k+1} \big[(k+1)!\big]^2 \\

&= \text{RHS} \\

\end{aligned} \\ \)

\( \therefore (2k+2)! > 2^{k+1} \big[(k+1)!\big]^2 \text{ for } \ge 2.\)

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