91Ó°ÊÓ

First, we need to check if the inequality holds for the base case when \(n = 1\). By plugging in n=1, we get: \[1! \leq 2^{-1}(1+1)^{1}\] \[1 \leq \frac{1}{2}(2)\] \[1 \leq 1\] The inequality holds, and since equality holds only when \(n=1\), our base case is proven.

Now we assume that the inequality holds for some \(n=k\), where \(k \in \mathbb{N}\). That means, we assume: \[k! \leq 2^{-k}(k+1)^{k}\]

Using our induction hypothesis, we now need to show that the inequality holds for \(n=k+1\). We have: \[(k+1)! \leq 2^{-(k+1)}[(k+1)+1]^{k+1}\] Let us start with the left-hand side of the inequality and see if we can show that it is less than or equal to the right-hand side: \[(k+1)! = (k+1)k!\] By using our induction hypothesis on \(k!\), we have: \[(k+1)! = (k+1)(2^{-k}(k+1)^{k})\] Now we need to show this expression is less than or equal to the one in the inequality for \(n=k+1\): \[(k+1)(2^{-k}(k+1)^{k}) \leq 2^{-(k+1)}[(k+1)+1]^{k+1}\] Divide both sides by \((k+1)\) to simplify: \[2^{-k}(k+1)^{k} \leq 2^{-(k+1)}(k+2)^{k+1}\] Divide both sides by \((k+1)^k\) to further simplify: \[2^{-k} \leq 2^{-(k+1)}\frac{(k+2)^{k+1}}{(k+1)^{k}}\] Now multiply both sides of the equation by \(2^{(k+1)}\) to cancel the 2 in the exponents: \[2 \leq \frac{(k+2)^{k+1}}{(k+1)^{k}}\] We can rewrite the right-hand side of the inequality as: \[\frac{(k+2)(k+2)^k}{(k+1)^k}\] Since \((k+2)>(k+1)\), we can rewrite the right-hand side as: \[\frac{(k+1)(k+2)^k}{(k+1)^k}\] This simplifies to: \[\frac{(k+2)^k}{(k+1)^{k-1}}\] Since \(k+2 \geq k+1\), every term in the numerator is greater than or equal to the corresponding term in the denominator, so we have: \[\frac{(k+2)^k}{(k+1)^{k-1}} \geq 2\] Thus, we've shown that: \[(k+1)! \leq 2^{-(k+1)}[(k+1)+1]^{k+1}\] By induction, the inequality holds for all natural numbers, and equality only holds when \(n=1\).