# Mathematical Induction Inequality Proof with Factorials

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