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I started with the base case that the statement is true when $n=4, 4!\geq 4^2$ because $24 \geq 16$ which is true.

For my second step I assumed that the statement is true when $n = k.$

$k! > k^ 2$ For each $k \geq 4$.

On the last step, I need to show that

$(k+1) ! \geq (k+1)^2$

I tried this

$(k+1)!=(k+1)k! = kk!+k! >=kk^2 +k^2$

This is where am stuck. I don't know where to go from here. I would appreciate any help.

amWhy
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user430638
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2 Answers2

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Factor and use the induction hypothesis to get to:

$(k+1)! = (k+1)\cdot k! \geq (k+1)\cdot k^2$

You should be able to follow this so far. Now... use the fact that $k\geq 4$ to recognize that $k+1\geq 3$. (Yes, this is a looser inequality and not as strict, but that is fine... we only needed it to be bigger than $3$, the extra tightness that we could have gotten by saying $k+1\geq 5$ is unnecessary)

$(k+1)\cdot k^2 \geq 3\cdot k^2 = k^2+k^2+k^2$

Now, use again the fact that $k\geq 3$ to see that

$k^2+k^2+k^2\geq k^2+2k+1$ which is precisely equal to the expansion of $(k+1)^2$

JMoravitz
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  • You only need that $k^2\geq k+1$ for $k\geq 2$ to get the result from the first inequality to the result. – Thomas Andrews Feb 06 '20 at 17:35
  • Can you please further explain the last step .. how did u get to the k^2+2k+1 part – user430638 Feb 06 '20 at 17:50
  • @user430638 Looking at the second term, we have $k^2=k\cdot k \geq 2\cdot k$. since again $k\geq 4\geq 2$. So, we can replace something bigger with something smaller in the inequality... $k^2+\color{blue}{k^2}+k^2\geq k^2 + \color{blue}{2k}+k^2$. Similarly, $k^2 = k\cdot k \geq 1\cdot k \geq 1\cdot 1 = 1$ so $k^2\geq 1$... so we have $k^2+\color{blue}{k^2}+\color{green}{k^2}\geq k^2+\color{blue}{2k}+\color{green}{1}$ – JMoravitz Feb 06 '20 at 17:55
  • @user430638 as for why I knew to try to aim for $k^2+2k+1$ in the first place? That is because it is common knowledge that $(k+1)^2 = k^2+2k+1$... seen easily by 'FOIL'ing or by binomial expansion, or however you prefer to explain it. – JMoravitz Feb 06 '20 at 17:56
  • @JMoravitz thank u so much! – user430638 Feb 06 '20 at 18:05
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Here’s a proof not using induction: by dividing both sides by $n$, this is equivalent to proving $$k! \geq (k+1)$$ for all $k \geq 3$. So if we find $k+1$ distinct permutations of the set $\{1,..,k\}$ we are done. First, we can clearly find $k$ permutations, since for $i$ ranging from $1$ to $k$, put $i$ in the first slot of the permutation and permute the rest of the $k-1$ numbers in any way you like and call this permutation $\pi_i$. Then the $\pi_1$,..,$\pi_k$ are $k$ distinct permutations of $\{1,..,k\}$.To find the $k+1$th permutation just look at $\pi_1$ and swap two numbers to the right of $1$ (this is possible because $k \geq 3$!).

M. Van
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